手机数字小键盘问题
给定移动数字键盘。您只能按下当前按钮的上、左、右或下按钮。不允许按下底部行的角按钮(即*和#)。
给定一个数 N，找出给定长度的可能数的个数。 例如: 对于 N=1，可能的数的数目是 10 (0，1，2，3，…。，9) 对于 N=2，可能的数字数将是 36 可能的数字:00，08 11，12，14 22，21，23，25 等等。 如果从 0 开始，有效数字将是 00，08(计数:2) 如果从 1 开始，有效数字将是 11，12，14(计数:3) 如果从 2 开始，有效数字将是 22，21，23，25(计数:4) 如果从 3 开始，有效数字将是 33，32，36(计数:3) 如果从 4 开始，有效数字将是 44，41，
N = 1 是平凡的情况，可能数字的个数是 10 (0，1，2，3，…，9) 对于 N > 1，我们需要从某个按钮开始，然后移动到四个方向(上、左、右或下)中的任何一个，这将导致一个有效的按钮(不应该转到*，#)。继续这样做，直到获得 N 个长度数(深度优先遍历)。
递归解决方案: 移动键盘是一个 4X3 的矩形网格(4 行 3 列) 假设 Count(i，j，N)代表从位置(I，j)开始的 N 个长度数字的计数
If N = 1
Count(i, j, N) = 10
Else
Count(i, j, N) = Sum of all Count(r, c, N-1) where (r, c) is new
position after valid move of length 1 from current
position (i, j)
下面是上述递归公式的实现。
C++
// A Naive Recursive C program to count number of possible numbers
// of given length
#include <stdio.h>
// left, up, right, down move from current location
int row[] = {0, 0, -1, 0, 1};
int col[] = {0, -1, 0, 1, 0};
// Returns count of numbers of length n starting from key position
// (i, j) in a numeric keyboard.
int getCountUtil(char keypad[][3], int i, int j, int n)
{
if (keypad == NULL || n <= 0)
return 0;
// From a given key, only one number is possible of length 1
if (n == 1)
return 1;
int k=0, move=0, ro=0, co=0, totalCount = 0;
// move left, up, right, down from current location and if
// new location is valid, then get number count of length
// (n-1) from that new position and add in count obtained so far
for (move=0; move<5; move++)
{
ro = i + row[move];
co = j + col[move];
if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
keypad[ro][co] != '*' && keypad[ro][co] != '#')
{
totalCount += getCountUtil(keypad, ro, co, n-1);
}
}
return totalCount;
}
// Return count of all possible numbers of length n
// in a given numeric keyboard
int getCount(char keypad[][3], int n)
{
// Base cases
if (keypad == NULL || n <= 0)
return 0;
if (n == 1)
return 10;
int i=0, j=0, totalCount = 0;
for (i=0; i<4; i++) // Loop on keypad row
{
for (j=0; j<3; j++) // Loop on keypad column
{
// Process for 0 to 9 digits
if (keypad[i][j] != '*' && keypad[i][j] != '#')
{
// Get count when number is starting from key
// position (i, j) and add in count obtained so far
totalCount += getCountUtil(keypad, i, j, n);
}
}
}
return totalCount;
}
// Driver program to test above function
int main(int argc, char *argv[])
{
char keypad[4][3] = {{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}};
printf("Count for numbers of length %d: %dn", 1, getCount(keypad, 1));
printf("Count for numbers of length %d: %dn", 2, getCount(keypad, 2));
printf("Count for numbers of length %d: %dn", 3, getCount(keypad, 3));
printf("Count for numbers of length %d: %dn", 4, getCount(keypad, 4));
printf("Count for numbers of length %d: %dn", 5, getCount(keypad, 5));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// A Naive Recursive Java program
// to count number of possible
// numbers of given length
class GfG
{
// left, up, right, down
// move from current location
static int row[] = {0, 0, -1, 0, 1};
static int col[] = {0, -1, 0, 1, 0};
// Returns count of numbers of length
// n starting from key position
// (i, j) in a numeric keyboard.
static int getCountUtil(char keypad[][],
int i, int j, int n)
{
if (keypad == null || n <= 0)
return 0;
// From a given key, only one
// number is possible of length 1
if (n == 1)
return 1;
int k = 0, move = 0, ro = 0, co = 0, totalCount = 0;
// move left, up, right, down
// from current location and if
// new location is valid, then
// get number count of length
// (n-1) from that new position
// and add in count obtained so far
for (move=0; move<5; move++)
{
ro = i + row[move];
co = j + col[move];
if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
keypad[ro][co] != '*' && keypad[ro][co] != '#')
{
totalCount += getCountUtil(keypad, ro, co, n - 1);
}
}
return totalCount;
}
// Return count of all possible numbers of length n
// in a given numeric keyboard
static int getCount(char keypad[][], int n)
{
// Base cases
if (keypad == null || n <= 0)
return 0;
if (n == 1)
return 10;
int i = 0, j = 0, totalCount = 0;
for (i = 0; i < 4; i++) // Loop on keypad row
{
for (j=0; j<3; j++) // Loop on keypad column
{
// Process for 0 to 9 digits
if (keypad[i][j] != '*' && keypad[i][j] != '#')
{
// Get count when number is starting from key
// position (i, j) and add in count obtained so far
totalCount += getCountUtil(keypad, i, j, n);
}
}
}
return totalCount;
}
// Driver code
public static void main(String[] args)
{
char keypad[][] = {{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}};
System.out.printf("Count for numbers of"+
" length %d: %d", 1, getCount(keypad, 1));
System.out.printf("\nCount for numbers of" +
"length %d: %d", 2, getCount(keypad, 2));
System.out.printf("\nCount for numbers of" +
"length %d: %d", 3, getCount(keypad, 3));
System.out.printf("\nCount for numbers of" +
"length %d: %d", 4, getCount(keypad, 4));
System.out.printf("\nCount for numbers of" +
"length %d: %d", 5, getCount(keypad, 5));
}
}
// This code has been contributed by 29AjayKumar
Python 3
# A Naive Recursive Python program to count number of possible numbers
# of given length
# left, up, right, down move from current location
row = [0, 0, -1, 0, 1]
col = [0, -1, 0, 1, 0]
# Returns count of numbers of length n starting from key position
# (i, j) in a numeric keyboard.
def getCountUtil(keypad, i, j, n):
if (keypad == None or n <= 0):
return 0
# From a given key, only one number is possible of length 1
if (n == 1):
return 1
k = 0
move = 0
ro = 0
co = 0
totalCount = 0
# move left, up, right, down from current location and if
# new location is valid, then get number count of length
# (n-1) from that new position and add in count obtained so far
for move in range(5):
ro = i + row[move]
co = j + col[move]
if (ro >= 0 and ro <= 3 and co >= 0 and co <= 2 and
keypad[ro][co] != '*' and keypad[ro][co] != '#'):
totalCount += getCountUtil(keypad, ro, co, n - 1)
return totalCount
# Return count of all possible numbers of length n
# in a given numeric keyboard
def getCount(keypad, n):
# Base cases
if (keypad == None or n <= 0):
return 0
if (n == 1):
return 10
i = 0
j = 0
totalCount = 0
for i in range(4): # Loop on keypad row
for j in range(3): # Loop on keypad column
# Process for 0 to 9 digits
if (keypad[i][j] != '*' and keypad[i][j] != '#'):
# Get count when number is starting from key
# position (i, j) and add in count obtained so far
totalCount += getCountUtil(keypad, i, j, n)
return totalCount
# Driver code
keypad = [['1', '2', '3'],
['4', '5', '6'],
['7', '8', '9'],
['*', '0', '#']]
print("Count for numbers of length 1:", getCount(keypad, 1))
print("Count for numbers of length 2:", getCount(keypad, 2))
print("Count for numbers of length 3:", getCount(keypad, 3))
print("Count for numbers of length 4:", getCount(keypad, 4))
print("Count for numbers of length 5:", getCount(keypad, 5))
# This code is contributed by subhammahato348
C
// A Naive Recursive C# program
// to count number of possible
// numbers of given length
using System;
class GfG
{
// left, up, right, down
// move from current location
static int []row = {0, 0, -1, 0, 1};
static int []col = {0, -1, 0, 1, 0};
// Returns count of numbers of length
// n starting from key position
// (i, j) in a numeric keyboard.
static int getCountUtil(char [,]keypad,
int i, int j, int n)
{
if (keypad == null || n <= 0)
return 0;
// From a given key, only one
// number is possible of length 1
if (n == 1)
return 1;
int k = 0, move = 0, ro = 0, co = 0, totalCount = 0;
// move left, up, right, down
// from current location and if
// new location is valid, then
// get number count of length
// (n-1) from that new position
// and add in count obtained so far
for (move=0; move<5; move++)
{
ro = i + row[move];
co = j + col[move];
if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
keypad[ro,co] != '*' && keypad[ro,co] != '#')
{
totalCount += getCountUtil(keypad, ro, co, n - 1);
}
}
return totalCount;
}
// Return count of all possible numbers of length n
// in a given numeric keyboard
static int getCount(char [,]keypad, int n)
{
// Base cases
if (keypad == null || n <= 0)
return 0;
if (n == 1)
return 10;
int i = 0, j = 0, totalCount = 0;
for (i = 0; i < 4; i++) // Loop on keypad row
{
for (j = 0; j < 3; j++) // Loop on keypad column
{
// Process for 0 to 9 digits
if (keypad[i, j] != '*' && keypad[i, j] != '#')
{
// Get count when number is starting from key
// position (i, j) and add in count obtained so far
totalCount += getCountUtil(keypad, i, j, n);
}
}
}
return totalCount;
}
// Driver code
public static void Main()
{
char [,]keypad = {{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}};
Console.Write("Count for numbers of"+
" length {0}: {1}", 1, getCount(keypad, 1));
Console.Write("\nCount for numbers of" +
"length {0}: {1}", 2, getCount(keypad, 2));
Console.Write("\nCount for numbers of" +
"length {0}: {1}", 3, getCount(keypad, 3));
Console.Write("\nCount for numbers of" +
"length {0}: {1}", 4, getCount(keypad, 4));
Console.Write("\nCount for numbers of" +
"length {0}: {1}", 5, getCount(keypad, 5));
}
}
/* This code contributed by PrinciRaj1992 */
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// A Naive Recursive PHP program
// to count number of possible
// numbers of given length
// left, up, right, down
// move from current location
// static $row = array(0, 0, -1, 0, 1);
//static $col = array(0, -1, 0, 1, 0);
// Returns count of numbers of length
// n starting from key position
// (i, j) in a numeric keyboard.
function getCountUtil($keypad,
$i, $j, $n)
{
static $row= array(0,0,-1,0,1);
static $col= array(0,-1,0,1,0);
if ($keypad == null || $n <= 0)
return 0;
// From a given key, only one
// number is possible of length 1
if ($n == 1)
return 1;
$k = 0; $move = 0; $ro = 0; $co = 0; $totalCount = 0;
// move left, up, right, down
// from current location and if
// new location is valid, then
// get number count of length
// (n-1) from that new position
// and add in count obtained so far
for ($move = 0; $move < 5; $move++)
{
$ro = $i + $row[$move];
$co = $j + $col[$move];
if ($ro >= 0 && $ro <= 3 && $co >=0 && $co <= 2 &&
$keypad[$ro][$co] != '*' && $keypad[$ro][$co] != '#')
{
$totalCount += getCountUtil($keypad, $ro, $co, $n - 1);
}
}
return $totalCount;
}
// Return count of all possible numbers of length n
// in a given numeric keyboard
function getCount($keypad, $n)
{
// Base cases
if ($keypad == null || $n <= 0)
return 0;
if ($n == 1)
return 10;
$i = 0; $j = 0; $totalCount = 0;
for ($i = 0; $i < 4; $i++) // Loop on keypad row
{
for ($j = 0; $j < 3; $j++) // Loop on keypad column
{
// Process for 0 to 9 digits
if ($keypad[$i][$j] != '*' && $keypad[$i][$j] != '#')
{
// Get count when number is starting from key
// position (i, j) and add in count obtained so far
$totalCount += getCountUtil($keypad, $i, $j, $n);
}
}
}
return $totalCount;
}
// Driver code
{
$keypad = array(array('1','2','3'),
array('4','5','6'),
array('7','8','9'),
array('*','0','#'));
echo("Count for numbers of"." length". getCount($keypad, 1));
echo("\nCount for numbers of" .
" length ". getCount($keypad, 2));
echo("\nCount for numbers of" .
" length ".getCount($keypad, 3));
echo("\nCount for numbers of" .
" length ".getCount($keypad, 4));
echo("\nCount for numbers of" .
" length ".getCount($keypad, 5));
}
// This code has been contributed by Code_Mech.
java 描述语言
<script>
// A Naive Recursive Javascript program
// to count number of possible
// numbers of given length
// left, up, right, down
// move from current location
let row=[0, 0, -1, 0, 1];
let col=[0, -1, 0, 1, 0];
// Returns count of numbers of length
// n starting from key position
// (i, j) in a numeric keyboard.
function getCountUtil(keypad,i,j,n)
{
if (keypad == null || n <= 0)
{return 0;}
// From a given key, only one
// number is possible of length 1
if (n == 1)
return 1;
let k = 0, move = 0, ro = 0, co = 0, totalCount = 0;
// move left, up, right, down
// from current location and if
// new location is valid, then
// get number count of length
// (n-1) from that new position
// and add in count obtained so far
for (move=0; move<5; move++)
{
ro = i + row[move];
co = j + col[move];
if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
keypad[ro][co] != '*' && keypad[ro][co] != '#')
{
totalCount += getCountUtil(keypad, ro, co, n - 1);
}
}
return totalCount;
}
// Return count of all possible numbers of length n
// in a given numeric keyboard
function getCount(keypad,n)
{
// Base cases
if (keypad == null || n <= 0)
return 0;
if (n == 1)
return 10;
let i = 0, j = 0, totalCount = 0;
for (i = 0; i < 4; i++) // Loop on keypad row
{
for (j=0; j<3; j++) // Loop on keypad column
{
// Process for 0 to 9 digits
if (keypad[i][j] != '*' && keypad[i][j] != '#')
{
// Get count when number is starting from key
// position (i, j) and add in count obtained so far
totalCount += getCountUtil(keypad, i, j, n);
}
}
}
return totalCount;
}
// Driver code
let keypad=[['1','2','3'],['4','5','6'],['7','8','9'],['*','0','#']];
document.write("Count for numbers of"+
" length ", 1,": ", getCount(keypad, 1));
document.write("<br>Count for numbers of" +
"length ", 2,": ", getCount(keypad, 2));
document.write("<br>Count for numbers of" +
"length ", 3,": ", getCount(keypad, 3));
document.write("<br>Count for numbers of" +
"length ", 4,": ", getCount(keypad, 4));
document.write("<br>Count for numbers of" +
"length ", 5,": ", getCount(keypad, 5));
// This code is contributed by avanitrachhadiya2155
</script>
输出:
Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062
动态规划 在较小的路径上有很多重复的遍历(较小 N 的遍历)来寻找所有可能的较长路径(较大 N 的遍历)。例如，请参见下面两个图表。在这个遍历中，对于从两个起始位置(按钮‘4’和‘8’)开始的 N = 4，我们可以看到对于 N = 2 的重复遍历很少(例如 4 - > 1，6 - > 3，8 - > 9，8 - > 7 等)。
由于该问题具有两个性质:最优子结构和重叠子问题，因此可以使用动态规划来有效地解决该问题。
下面是动态编程实现的程序。
C++
// A Dynamic Programming based C program to count number of
// possible numbers of given length
#include <stdio.h>
// Return count of all possible numbers of length n
// in a given numeric keyboard
int getCount(char keypad[][3], int n)
{
if(keypad == NULL || n <= 0)
return 0;
if(n == 1)
return 10;
// left, up, right, down move from current location
int row[] = {0, 0, -1, 0, 1};
int col[] = {0, -1, 0, 1, 0};
// taking n+1 for simplicity - count[i][j] will store
// number count starting with digit i and length j
int count[10][n+1];
int i=0, j=0, k=0, move=0, ro=0, co=0, num = 0;
int nextNum=0, totalCount = 0;
// count numbers starting with digit i and of lengths 0 and 1
for (i=0; i<=9; i++)
{
count[i][0] = 0;
count[i][1] = 1;
}
// Bottom up - Get number count of length 2, 3, 4, ... , n
for (k=2; k<=n; k++)
{
for (i=0; i<4; i++) // Loop on keypad row
{
for (j=0; j<3; j++) // Loop on keypad column
{
// Process for 0 to 9 digits
if (keypad[i][j] != '*' && keypad[i][j] != '#')
{
// Here we are counting the numbers starting with
// digit keypad[i][j] and of length k keypad[i][j]
// will become 1st digit, and we need to look for
// (k-1) more digits
num = keypad[i][j] - '0';
count[num][k] = 0;
// move left, up, right, down from current location
// and if new location is valid, then get number
// count of length (k-1) from that new digit and
// add in count we found so far
for (move=0; move<5; move++)
{
ro = i + row[move];
co = j + col[move];
if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
keypad[ro][co] != '*' && keypad[ro][co] != '#')
{
nextNum = keypad[ro][co] - '0';
count[num][k] += count[nextNum][k-1];
}
}
}
}
}
}
// Get count of all possible numbers of length "n" starting
// with digit 0, 1, 2, ..., 9
totalCount = 0;
for (i=0; i<=9; i++)
totalCount += count[i][n];
return totalCount;
}
// Driver program to test above function
int main(int argc, char *argv[])
{
char keypad[4][3] = {{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}};
printf("Count for numbers of length %d: %dn", 1, getCount(keypad, 1));
printf("Count for numbers of length %d: %dn", 2, getCount(keypad, 2));
printf("Count for numbers of length %d: %dn", 3, getCount(keypad, 3));
printf("Count for numbers of length %d: %dn", 4, getCount(keypad, 4));
printf("Count for numbers of length %d: %dn", 5, getCount(keypad, 5));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// A Dynamic Programming based Java program to
// count number of possible numbers of given length
class GFG
{
// Return count of all possible numbers of length n
// in a given numeric keyboard
static int getCount(char keypad[][], int n)
{
if(keypad == null || n <= 0)
return 0;
if(n == 1)
return 10;
// left, up, right, down move from current location
int row[] = {0, 0, -1, 0, 1};
int col[] = {0, -1, 0, 1, 0};
// taking n+1 for simplicity - count[i][j] will store
// number count starting with digit i and length j
int [][]count = new int[10][n + 1];
int i = 0, j = 0, k = 0, move = 0,
ro = 0, co = 0, num = 0;
int nextNum = 0, totalCount = 0;
// count numbers starting with digit i
// and of lengths 0 and 1
for (i = 0; i <= 9; i++)
{
count[i][0] = 0;
count[i][1] = 1;
}
// Bottom up - Get number count of length 2, 3, 4, ... , n
for (k = 2; k <= n; k++)
{
for (i = 0; i < 4; i++) // Loop on keypad row
{
for (j = 0; j < 3; j++) // Loop on keypad column
{
// Process for 0 to 9 digits
if (keypad[i][j] != '*' &&
keypad[i][j] != '#')
{
// Here we are counting the numbers starting with
// digit keypad[i][j] and of length k keypad[i][j]
// will become 1st digit, and we need to look for
// (k-1) more digits
num = keypad[i][j] - '0';
count[num][k] = 0;
// move left, up, right, down from current location
// and if new location is valid, then get number
// count of length (k-1) from that new digit and
// add in count we found so far
for (move = 0; move < 5; move++)
{
ro = i + row[move];
co = j + col[move];
if (ro >= 0 && ro <= 3 && co >= 0 &&
co <= 2 && keypad[ro][co] != '*' &&
keypad[ro][co] != '#')
{
nextNum = keypad[ro][co] - '0';
count[num][k] += count[nextNum][k - 1];
}
}
}
}
}
}
// Get count of all possible numbers of length "n"
// starting with digit 0, 1, 2, ..., 9
totalCount = 0;
for (i = 0; i <= 9; i++)
totalCount += count[i][n];
return totalCount;
}
// Driver Code
public static void main(String[] args)
{
char keypad[][] = {{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}};
System.out.printf("Count for numbers of length %d: %d\n", 1,
getCount(keypad, 1));
System.out.printf("Count for numbers of length %d: %d\n", 2,
getCount(keypad, 2));
System.out.printf("Count for numbers of length %d: %d\n", 3,
getCount(keypad, 3));
System.out.printf("Count for numbers of length %d: %d\n", 4,
getCount(keypad, 4));
System.out.printf("Count for numbers of length %d: %d\n", 5,
getCount(keypad, 5));
}
}
// This code is contributed by Rajput-Ji
Python 3
# A Dynamic Programming based C program to count number of
# possible numbers of given length
# Return count of all possible numbers of length n
# in a given numeric keyboard
def getCount(keypad, n):
if (keypad == None or n <= 0):
return 0
if (n == 1):
return 10
# left, up, right, down move from current location
row = [0, 0, -1, 0, 1]
col = [0, -1, 0, 1, 0]
# taking n+1 for simplicity - count[i][j] will store
# number count starting with digit i and length j
# count[10][n+1]
count = [[0]*(n + 1)]*10
i = 0
j = 0
k = 0
move = 0
ro = 0
co = 0
num = 0
nextNum = 0
totalCount = 0
# count numbers starting with
# digit i and of lengths 0 and 1
for i in range(10):
count[i][0] = 0
count[i][1] = 1
# Bottom up - Get number
# count of length 2, 3, 4, ... , n
for k in range(2, n + 1):
for i in range(4): # Loop on keypad row
for j in range(3): # Loop on keypad column
# Process for 0 to 9 digits
if (keypad[i][j] != '*' and keypad[i][j] != '#'):
# Here we are counting the numbers starting with
# digit keypad[i][j] and of length k keypad[i][j]
# will become 1st digit, and we need to look for
# (k-1) more digits
num = ord(keypad[i][j]) - 48
count[num][k] = 0
# move left, up, right, down from current location
# and if new location is valid, then get number
# count of length (k-1) from that new digit and
# add in count we found so far
for move in range(5):
ro = i + row[move]
co = j + col[move]
if (ro >= 0 and ro <= 3 and co >= 0 and co <= 2 and
keypad[ro][co] != '*' and keypad[ro][co] != '#'):
nextNum = ord(keypad[ro][co]) - 48
count[num][k] += count[nextNum][k - 1]
# Get count of all possible numbers of length "n" starting
# with digit 0, 1, 2, ..., 9
totalCount = 0
for i in range(10):
totalCount += count[i][n]
return totalCount
# Driver code
if __name__ == "__main__":
keypad = [['1','2','3'],
['4','5','6'],
['7','8','9'],
['*','0','#']]
print("Count for numbers of length", 1, ":", getCount(keypad, 1))
print("Count for numbers of length", 2, ":", getCount(keypad, 2))
print("Count for numbers of length", 3, ":", getCount(keypad, 3))
print("Count for numbers of length", 4, ":", getCount(keypad, 4))
print("Count for numbers of length", 5, ":", getCount(keypad, 5))
# This code is contributed by subhammahato348
C
// A Dynamic Programming based C# program to
// count number of possible numbers of given Length
using System;
class GFG
{
// Return count of all possible numbers of Length n
// in a given numeric keyboard
static int getCount(char [,]keypad, int n)
{
if(keypad == null || n <= 0)
return 0;
if(n == 1)
return 10;
// left, up, right, down move
// from current location
int []row = {0, 0, -1, 0, 1};
int []col = {0, -1, 0, 1, 0};
// taking n+1 for simplicity - count[i,j]
// will store number count starting with
// digit i and.Length j
int [,]count = new int[10,n + 1];
int i = 0, j = 0, k = 0, move = 0,
ro = 0, co = 0, num = 0;
int nextNum = 0, totalCount = 0;
// count numbers starting with digit i
// and of.Lengths 0 and 1
for (i = 0; i <= 9; i++)
{
count[i, 0] = 0;
count[i, 1] = 1;
}
// Bottom up - Get number count of
// Length 2, 3, 4, ... , n
for (k = 2; k <= n; k++)
{
for (i = 0; i < 4; i++) // Loop on keypad row
{
for (j = 0; j < 3; j++) // Loop on keypad column
{
// Process for 0 to 9 digits
if (keypad[i, j] != '*' &&
keypad[i, j] != '#')
{
// Here we are counting the numbers starting with
// digit keypad[i,j] and of.Length k keypad[i,j]
// will become 1st digit, and we need to look for
// (k-1) more digits
num = keypad[i, j] - '0';
count[num, k] = 0;
// move left, up, right, down from current location
// and if new location is valid, then get number
// count of.Length (k-1) from that new digit and
//.Add in count we found so far
for (move = 0; move < 5; move++)
{
ro = i + row[move];
co = j + col[move];
if (ro >= 0 && ro <= 3 && co >= 0 &&
co <= 2 && keypad[ro, co] != '*' &&
keypad[ro, co] != '#')
{
nextNum = keypad[ro, co] - '0';
count[num, k] += count[nextNum, k - 1];
}
}
}
}
}
}
// Get count of all possible numbers of.Length "n"
// starting with digit 0, 1, 2, ..., 9
totalCount = 0;
for (i = 0; i <= 9; i++)
totalCount += count[i, n];
return totalCount;
}
// Driver Code
public static void Main(String[] args)
{
char [,]keypad = {{'1', '2', '3'},
{'4', '5', '6'},
{'7', '8', '9'},
{'*', '0', '#'}};
Console.Write("Count for numbers of.Length {0}: {1}\n", 1,
getCount(keypad, 1));
Console.Write("Count for numbers of.Length {0}: {1}\n", 2,
getCount(keypad, 2));
Console.Write("Count for numbers of.Length {0}: {1}\n", 3,
getCount(keypad, 3));
Console.Write("Count for numbers of.Length {0}: {1}\n", 4,
getCount(keypad, 4));
Console.Write("Count for numbers of.Length {0}: {1}\n", 5,
getCount(keypad, 5));
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// A Dynamic Programming based Javascript program to
// count number of possible numbers of given length
// Return count of all possible numbers of length n
// in a given numeric keyboard
function getCount(keypad, n)
{
if (keypad == null || n <= 0)
return 0;
if (n == 1)
return 10;
// left, up, right, down move
// from current location
let row = [ 0, 0, -1, 0, 1 ];
let col = [ 0, -1, 0, 1, 0 ];
// Taking n+1 for simplicity - count[i][j]
// will store number count starting with
// digit i and length j
let count = new Array(10);
for(let i = 0; i < 10; i++)
{
count[i] = new Array(n + 1);
for(let j = 0; j < n + 1; j++)
{
count[i][j] = 0;
}
}
let i = 0, j = 0, k = 0, move = 0,
ro = 0, co = 0, num = 0;
let nextNum = 0, totalCount = 0;
// count numbers starting with digit i
// and of lengths 0 and 1
for(i = 0; i <= 9; i++)
{
count[i][0] = 0;
count[i][1] = 1;
}
// Bottom up - Get number count
// of length 2, 3, 4, ... , n
for(k = 2; k <= n; k++)
{
// Loop on keypad row
for(i = 0; i < 4; i++)
{
// Loop on keypad column
for(j = 0; j < 3; j++)
{
// Process for 0 to 9 digits
if (keypad[i][j] != '*' &&
keypad[i][j] != '#')
{
// Here we are counting the numbers starting with
// digit keypad[i][j] and of length k keypad[i][j]
// will become 1st digit, and we need to look for
// (k-1) more digits
num = keypad[i][j].charCodeAt(0) -
'0'.charCodeAt(0);
count[num][k] = 0;
// Move left, up, right, down from current location
// and if new location is valid, then get number
// count of length (k-1) from that new digit and
// add in count we found so far
for(move = 0; move < 5; move++)
{
ro = i + row[move];
co = j + col[move];
if (ro >= 0 && ro <= 3 && co >= 0 &&
co <= 2 && keypad[ro][co] != '*' &&
keypad[ro][co] != '#')
{
nextNum = keypad[ro][co].charCodeAt(0) -
'0'.charCodeAt(0);
count[num][k] += count[nextNum][k - 1];
}
}
}
}
}
}
// Get count of all possible numbers of length "n"
// starting with digit 0, 1, 2, ..., 9
totalCount = 0;
for(i = 0; i <= 9; i++)
totalCount += count[i][n];
return totalCount;
}
// Driver Code
let keypad = [ [ '1','2','3' ],
[ '4','5','6' ],
[ '7','8','9' ],
[ '*','0','#' ] ];
document.write("Count for numbers of length " +
1 + " : " + getCount(keypad, 1) + "<br>")
document.write("Count for numbers of length " +
2 + " : " + getCount(keypad, 2) + "<br>")
document.write("Count for numbers of length " +
3 + " : " + getCount(keypad, 3) + "<br>")
document.write("Count for numbers of length " +
4 + " : " + getCount(keypad, 4) + "<br>")
document.write("Count for numbers of length " +
5 + " : " + getCount(keypad, 5) + "<br>")
// This code is contributed by rag2127
</script>
输出:
Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062
一个空间优化的解决方案: 上述动态规划方法也在 O(n)个时间内运行，并且需要 O(n)个辅助空间，因为只有一个用于循环运行 n 次，其他用于循环运行恒定时间。我们可以看到，第 n 次迭代只需要来自第(n-1)次迭代的数据，所以我们不需要保留来自旧迭代的数据。我们可以有一个空间高效的动态编程方法，只有两个大小为 10 的数组。感谢 Nik 提出这个解决方案。
C++
// A Space Optimized C++ program to count number of possible numbers
// of given length
#include <bits/stdc++.h>
using namespace std;
// Return count of all possible numbers of length n
// in a given numeric keyboard
int getCount(char keypad[][3], int n)
{
if (keypad == NULL || n <= 0)
return 0;
if (n == 1)
return 10;
// odd[i], even[i] arrays represent count of numbers starting
// with digit i for any length j
int odd[10], even[10];
int i = 0, j = 0, useOdd = 0, totalCount = 0;
for (i = 0; i <= 9; i++)
odd[i] = 1; // for j = 1
for (j = 2; j <= n; j++) // Bottom Up calculation from j = 2 to n
{
useOdd = 1 - useOdd;
// Here we are explicitly writing lines for each number 0
// to 9\. But it can always be written as DFS on 4X3 grid
// using row, column array valid moves
if (useOdd == 1)
{
even[0] = odd[0] + odd[8];
even[1] = odd[1] + odd[2] + odd[4];
even[2] = odd[2] + odd[1] + odd[3] + odd[5];
even[3] = odd[3] + odd[2] + odd[6];
even[4] = odd[4] + odd[1] + odd[5] + odd[7];
even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6];
even[6] = odd[6] + odd[3] + odd[5] + odd[9];
even[7] = odd[7] + odd[4] + odd[8];
even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9];
even[9] = odd[9] + odd[6] + odd[8];
}
else
{
odd[0] = even[0] + even[8];
odd[1] = even[1] + even[2] + even[4];
odd[2] = even[2] + even[1] + even[3] + even[5];
odd[3] = even[3] + even[2] + even[6];
odd[4] = even[4] + even[1] + even[5] + even[7];
odd[5] = even[5] + even[2] + even[4] + even[8] + even[6];
odd[6] = even[6] + even[3] + even[5] + even[9];
odd[7] = even[7] + even[4] + even[8];
odd[8] = even[8] + even[0] + even[5] + even[7] + even[9];
odd[9] = even[9] + even[6] + even[8];
}
}
// Get count of all possible numbers of length "n" starting
// with digit 0, 1, 2, ..., 9
totalCount = 0;
if (useOdd == 1)
{
for (i = 0; i <= 9; i++)
totalCount += even[i];
}
else
{
for (i = 0; i <= 9; i++)
totalCount += odd[i];
}
return totalCount;
}
// Driver program to test above function
int main()
{
char keypad[4][3] = {{'1', '2', '3'},
{'4', '5', '6'},
{'7', '8', '9'},
{'*', '0', '#'}};
cout << "Count for numbers of length 1: " << getCount(keypad, 1) << endl;
cout << "Count for numbers of length 2: " << getCount(keypad, 2) << endl;
cout << "Count for numbers of length 3: " << getCount(keypad, 3) << endl;
cout << "Count for numbers of length 4: " << getCount(keypad, 4) << endl;
cout << "Count for numbers of length 5: " << getCount(keypad, 5) << endl;
return 0;
}
//This code is contributed by Mayank Tyagi
C
// A Space Optimized C program to count number of possible numbers
// of given length
#include <stdio.h>
// Return count of all possible numbers of length n
// in a given numeric keyboard
int getCount(char keypad[][3], int n)
{
if(keypad == NULL || n <= 0)
return 0;
if(n == 1)
return 10;
// odd[i], even[i] arrays represent count of numbers starting
// with digit i for any length j
int odd[10], even[10];
int i = 0, j = 0, useOdd = 0, totalCount = 0;
for (i=0; i<=9; i++)
odd[i] = 1; // for j = 1
for (j=2; j<=n; j++) // Bottom Up calculation from j = 2 to n
{
useOdd = 1 - useOdd;
// Here we are explicitly writing lines for each number 0
// to 9\. But it can always be written as DFS on 4X3 grid
// using row, column array valid moves
if(useOdd == 1)
{
even[0] = odd[0] + odd[8];
even[1] = odd[1] + odd[2] + odd[4];
even[2] = odd[2] + odd[1] + odd[3] + odd[5];
even[3] = odd[3] + odd[2] + odd[6];
even[4] = odd[4] + odd[1] + odd[5] + odd[7];
even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6];
even[6] = odd[6] + odd[3] + odd[5] + odd[9];
even[7] = odd[7] + odd[4] + odd[8];
even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9];
even[9] = odd[9] + odd[6] + odd[8];
}
else
{
odd[0] = even[0] + even[8];
odd[1] = even[1] + even[2] + even[4];
odd[2] = even[2] + even[1] + even[3] + even[5];
odd[3] = even[3] + even[2] + even[6];
odd[4] = even[4] + even[1] + even[5] + even[7];
odd[5] = even[5] + even[2] + even[4] + even[8] + even[6];
odd[6] = even[6] + even[3] + even[5] + even[9];
odd[7] = even[7] + even[4] + even[8];
odd[8] = even[8] + even[0] + even[5] + even[7] + even[9];
odd[9] = even[9] + even[6] + even[8];
}
}
// Get count of all possible numbers of length "n" starting
// with digit 0, 1, 2, ..., 9
totalCount = 0;
if(useOdd == 1)
{
for (i=0; i<=9; i++)
totalCount += even[i];
}
else
{
for (i=0; i<=9; i++)
totalCount += odd[i];
}
return totalCount;
}
// Driver program to test above function
int main()
{
char keypad[4][3] = {{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}
};
printf("Count for numbers of length %d: %dn", 1, getCount(keypad, 1));
printf("Count for numbers of length %d: %dn", 2, getCount(keypad, 2));
printf("Count for numbers of length %d: %dn", 3, getCount(keypad, 3));
printf("Count for numbers of length %d: %dn", 4, getCount(keypad, 4));
printf("Count for numbers of length %d: %dn", 5, getCount(keypad, 5));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// A Space Optimized Java program to
// count number of possible numbers
// of given length
class GFG
{
// Return count of all possible numbers of
// length n in a given numeric keyboard
static int getCount(char keypad[][], int n)
{
if(keypad == null || n <= 0)
return 0;
if(n == 1)
return 10;
// odd[i], even[i] arrays represent count of
// numbers starting with digit i for any length j
int []odd = new int[10];
int []even = new int[10];
int i = 0, j = 0, useOdd = 0, totalCount = 0;
for (i = 0; i <= 9; i++)
odd[i] = 1; // for j = 1
// Bottom Up calculation from j = 2 to n
for (j = 2; j <= n; j++)
{
useOdd = 1 - useOdd;
// Here we are explicitly writing lines
// for each number 0 to 9\. But it can always be
// written as DFS on 4X3 grid using row,
// column array valid moves
if(useOdd == 1)
{
even[0] = odd[0] + odd[8];
even[1] = odd[1] + odd[2] + odd[4];
even[2] = odd[2] + odd[1] +
odd[3] + odd[5];
even[3] = odd[3] + odd[2] + odd[6];
even[4] = odd[4] + odd[1] +
odd[5] + odd[7];
even[5] = odd[5] + odd[2] + odd[4] +
odd[8] + odd[6];
even[6] = odd[6] + odd[3] +
odd[5] + odd[9];
even[7] = odd[7] + odd[4] + odd[8];
even[8] = odd[8] + odd[0] + odd[5] +
odd[7] + odd[9];
even[9] = odd[9] + odd[6] + odd[8];
}
else
{
odd[0] = even[0] + even[8];
odd[1] = even[1] + even[2] + even[4];
odd[2] = even[2] + even[1] +
even[3] + even[5];
odd[3] = even[3] + even[2] + even[6];
odd[4] = even[4] + even[1] +
even[5] + even[7];
odd[5] = even[5] + even[2] + even[4] +
even[8] + even[6];
odd[6] = even[6] + even[3] +
even[5] + even[9];
odd[7] = even[7] + even[4] + even[8];
odd[8] = even[8] + even[0] + even[5] +
even[7] + even[9];
odd[9] = even[9] + even[6] + even[8];
}
}
// Get count of all possible numbers of
// length "n" starting with digit 0, 1, 2, ..., 9
totalCount = 0;
if(useOdd == 1)
{
for (i = 0; i <= 9; i++)
totalCount += even[i];
}
else
{
for (i = 0; i <= 9; i++)
totalCount += odd[i];
}
return totalCount;
}
// Driver Code
public static void main(String[] args)
{
char keypad[][] = {{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}};
System.out.printf("Count for numbers of length %d: %d\n", 1,
getCount(keypad, 1));
System.out.printf("Count for numbers of length %d: %d\n", 2,
getCount(keypad, 2));
System.out.printf("Count for numbers of length %d: %d\n", 3,
getCount(keypad, 3));
System.out.printf("Count for numbers of length %d: %d\n", 4,
getCount(keypad, 4));
System.out.printf("Count for numbers of length %d: %d\n", 5,
getCount(keypad, 5));
}
}
// This code is contributed by PrinciRaj1992
Python 3
# A Space Optimized Python program to count
# number of possible numbers
# of given length
# Return count of all possible numbers
# of length n
# in a given numeric keyboard
def getCount(keypad, n):
if(not keypad or n <= 0):
return 0
if(n == 1):
return 10
# odd[i], even[i] arrays represent
# count of numbers starting
# with digit i for any length j
odd = [0]*10
even = [0]*10
i = 0
j = 0
useOdd = 0
totalCount = 0
for i in range(10):
odd[i] = 1 # for j = 1
for j in range(2,n+1): # Bottom Up calculation from j = 2 to n
useOdd = 1 - useOdd
# Here we are explicitly writing lines for each number 0
# to 9\. But it can always be written as DFS on 4X3 grid
# using row, column array valid moves
if(useOdd == 1):
even[0] = odd[0] + odd[8]
even[1] = odd[1] + odd[2] + odd[4]
even[2] = odd[2] + odd[1] + odd[3] + odd[5]
even[3] = odd[3] + odd[2] + odd[6]
even[4] = odd[4] + odd[1] + odd[5] + odd[7]
even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6]
even[6] = odd[6] + odd[3] + odd[5] + odd[9]
even[7] = odd[7] + odd[4] + odd[8]
even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9]
even[9] = odd[9] + odd[6] + odd[8]
else:
odd[0] = even[0] + even[8]
odd[1] = even[1] + even[2] + even[4]
odd[2] = even[2] + even[1] + even[3] + even[5]
odd[3] = even[3] + even[2] + even[6]
odd[4] = even[4] + even[1] + even[5] + even[7]
odd[5] = even[5] + even[2] + even[4] + even[8] + even[6]
odd[6] = even[6] + even[3] + even[5] + even[9]
odd[7] = even[7] + even[4] + even[8]
odd[8] = even[8] + even[0] + even[5] + even[7] + even[9]
odd[9] = even[9] + even[6] + even[8]
# Get count of all possible numbers of length "n" starting
# with digit 0, 1, 2, ..., 9
totalCount = 0
if(useOdd == 1):
for i in range(10):
totalCount += even[i]
else:
for i in range(10):
totalCount += odd[i]
return totalCount
# Driver program to test above function
if __name__ == "__main__":
keypad = [['1','2','3'],
['4','5','6'],
['7','8','9'],
['*','0','#']]
print("Count for numbers of length ",1,": ", getCount(keypad, 1))
print("Count for numbers of length ",2,": ", getCount(keypad, 2))
print("Count for numbers of length ",3,": ", getCount(keypad, 3))
print("Count for numbers of length ",4,": ", getCount(keypad, 4))
print("Count for numbers of length ",5,": ", getCount(keypad, 5))
# This code is contributed by
# ChitraNayal
C
// A Space Optimized C# program to
// count number of possible numbers
// of given length
using System;
class GFG
{
// Return count of all possible numbers of
// length n in a given numeric keyboard
static int getCount(char [,]keypad, int n)
{
if(keypad == null || n <= 0)
return 0;
if(n == 1)
return 10;
// odd[i], even[i] arrays represent count of
// numbers starting with digit i for any length j
int []odd = new int[10];
int []even = new int[10];
int i = 0, j = 0, useOdd = 0, totalCount = 0;
for (i = 0; i <= 9; i++)
odd[i] = 1; // for j = 1
// Bottom Up calculation from j = 2 to n
for (j = 2; j <= n; j++)
{
useOdd = 1 - useOdd;
// Here we are explicitly writing lines
// for each number 0 to 9\. But it can always be
// written as DFS on 4X3 grid using row,
// column array valid moves
if(useOdd == 1)
{
even[0] = odd[0] + odd[8];
even[1] = odd[1] + odd[2] + odd[4];
even[2] = odd[2] + odd[1] +
odd[3] + odd[5];
even[3] = odd[3] + odd[2] + odd[6];
even[4] = odd[4] + odd[1] +
odd[5] + odd[7];
even[5] = odd[5] + odd[2] + odd[4] +
odd[8] + odd[6];
even[6] = odd[6] + odd[3] +
odd[5] + odd[9];
even[7] = odd[7] + odd[4] + odd[8];
even[8] = odd[8] + odd[0] + odd[5] +
odd[7] + odd[9];
even[9] = odd[9] + odd[6] + odd[8];
}
else
{
odd[0] = even[0] + even[8];
odd[1] = even[1] + even[2] + even[4];
odd[2] = even[2] + even[1] +
even[3] + even[5];
odd[3] = even[3] + even[2] + even[6];
odd[4] = even[4] + even[1] +
even[5] + even[7];
odd[5] = even[5] + even[2] + even[4] +
even[8] + even[6];
odd[6] = even[6] + even[3] +
even[5] + even[9];
odd[7] = even[7] + even[4] + even[8];
odd[8] = even[8] + even[0] + even[5] +
even[7] + even[9];
odd[9] = even[9] + even[6] + even[8];
}
}
// Get count of all possible numbers of
// length "n" starting with digit 0, 1, 2, ..., 9
totalCount = 0;
if(useOdd == 1)
{
for (i = 0; i <= 9; i++)
totalCount += even[i];
}
else
{
for (i = 0; i <= 9; i++)
totalCount += odd[i];
}
return totalCount;
}
// Driver Code
public static void Main(String[] args)
{
char [,]keypad = {{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}};
Console.Write("Count for numbers of length {0}: {1}\n", 1,
getCount(keypad, 1));
Console.Write("Count for numbers of length {0}: {1}\n", 2,
getCount(keypad, 2));
Console.Write("Count for numbers of length {0}: {1}\n", 3,
getCount(keypad, 3));
Console.Write("Count for numbers of length {0}: {1}\n", 4,
getCount(keypad, 4));
Console.Write("Count for numbers of length {0}: {1}\n", 5,
getCount(keypad, 5));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// A Space Optimized javascript program to
// count number of possible numbers
// of given length
// Return count of all possible numbers of
// length n in a given numeric keyboard
function getCount(keypad , n)
{
if(keypad == null || n <= 0)
return 0;
if(n == 1)
return 10;
// odd[i], even[i] arrays represent count of
// numbers starting with digit i for any length j
var odd = Array.from({length: 10}, (_, i) => 0);
var even = Array.from({length: 10}, (_, i) => 0);
var i = 0, j = 0, useOdd = 0, totalCount = 0;
for (i = 0; i <= 9; i++)
odd[i] = 1; // for j = 1
// Bottom Up calculation from j = 2 to n
for (j = 2; j <= n; j++)
{
useOdd = 1 - useOdd;
// Here we are explicitly writing lines
// for each number 0 to 9\. But it can always be
// written as DFS on 4X3 grid using row,
// column array valid moves
if(useOdd == 1)
{
even[0] = odd[0] + odd[8];
even[1] = odd[1] + odd[2] + odd[4];
even[2] = odd[2] + odd[1] +
odd[3] + odd[5];
even[3] = odd[3] + odd[2] + odd[6];
even[4] = odd[4] + odd[1] +
odd[5] + odd[7];
even[5] = odd[5] + odd[2] + odd[4] +
odd[8] + odd[6];
even[6] = odd[6] + odd[3] +
odd[5] + odd[9];
even[7] = odd[7] + odd[4] + odd[8];
even[8] = odd[8] + odd[0] + odd[5] +
odd[7] + odd[9];
even[9] = odd[9] + odd[6] + odd[8];
}
else
{
odd[0] = even[0] + even[8];
odd[1] = even[1] + even[2] + even[4];
odd[2] = even[2] + even[1] +
even[3] + even[5];
odd[3] = even[3] + even[2] + even[6];
odd[4] = even[4] + even[1] +
even[5] + even[7];
odd[5] = even[5] + even[2] + even[4] +
even[8] + even[6];
odd[6] = even[6] + even[3] +
even[5] + even[9];
odd[7] = even[7] + even[4] + even[8];
odd[8] = even[8] + even[0] + even[5] +
even[7] + even[9];
odd[9] = even[9] + even[6] + even[8];
}
}
// Get count of all possible numbers of
// length "n" starting with digit 0, 1, 2, ..., 9
totalCount = 0;
if(useOdd == 1)
{
for (i = 0; i <= 9; i++)
totalCount += even[i];
}
else
{
for (i = 0; i <= 9; i++)
totalCount += odd[i];
}
return totalCount;
}
// Driver Code
var keypad = [['1','2','3'],
['4','5','6'],
['7','8','9'],
['*','0','#']];
document.write("Count for numbers of length "+ 1+": "+
getCount(keypad, 1));
document.write("<br>Count for numbers of length "+ 2+": "+
getCount(keypad, 2));
document.write("<br>Count for numbers of length "+ 3+": "+
getCount(keypad, 3));
document.write("<br>Count for numbers of length "+ 4+": "+
getCount(keypad, 4));
document.write("<br>Count for numbers of length "+ 5+": "+
getCount(keypad, 5));
// This code is contributed by 29AjayKumar
</script>
输出:
Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062
本文由阿努拉格·辛格供稿。如果你发现任何不正确的地方，或者你想分享更多关于上面讨论的话题的信息，请写评论。
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