从一个数字结束的第 K 个数字
给定一个整数 N ，任务是从一个整数 N 的末尾找到 K 第位。如果 K 第位不存在，则打印 -1 。
示例:
输入 : N = 2354，K = 2 输出 : 5
输入 : N = 1234，K = 1 输出 : 4
天真法:解决这个问题最简单的方法就是将整数 N 转换为字符串。按照以下步骤解决此问题:
- 如果 K 小于等于 0，则打印 -1 返回。
- 将 N 转换成字符串，储存在温度中。
- 如果 K 大于温度的大小，则打印 -1 ，否则从最后打印 K 字符。
下面是上述方法的实现:
C++
// c++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the kth digit
// from last in an integer n
void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0) {
cout << -1 << endl;
return;
}
// Convert integer into string
string temp = to_string(n);
// If k is greater than length of the
// string temp
if (k > temp.length()) {
cout << -1 << endl;
}
// Print the k digit from last
else {
cout << temp[temp.length() - k] - '0';
}
}
// Driver code
int main()
{
// Given Input
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
class GFG{
// Function to find the kth digit
// from last in an integer n
public static void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0)
{
System.out.println(-1);
return;
}
// Convert integer into string
String temp = Integer.toString(n);
// If k is greater than length of the
// string temp
if (k > temp.length())
{
System.out.println(-1);
}
// Print the k digit from last
else
{
int index = temp.length() - k;
int res = temp.charAt(index) - '0';
System.out.println(res);
}
}
// Driver code
public static void main(String[] args)
{
// Given Input
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
}
// This code is contributed by Potta Lokesh
Python 3
# Python3 program for the above approach
# Function to find the kth digit
# from last in an integer n
def kthDigitFromLast(n, k):
# If k is less than equal to 0
if (k <= 0):
print(-1)
return
# Convert integer into string
temp = str(n)
# If k is greater than length of the
# temp
if (k > len(temp)):
print(-1)
# Print the k digit from last
else:
print(ord(temp[len(temp) - k]) - ord('0'))
# Driver code
if __name__ == '__main__':
# Given Input
n = 2354
k = 2
# Function call
kthDigitFromLast(n, k)
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the kth digit
// from last in an integer n
static void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0)
{
Console.Write(-1);
return;
}
// Convert integer into string
string temp = n.ToString();
// If k is greater than length of the
// string temp
if (k > temp.Length)
{
Console.WriteLine(-1);
}
// Print the k digit from last
else
{
Console.Write(temp[temp.Length - k] - 48);
}
}
// Driver code
public static void Main()
{
// Given Input
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
}
// This code is contributed by SURENDRA_GANGWAR
java 描述语言
<script>
// JavaScript program for the above approach
// Function to find the kth digit
// from last in an integer n
function kthDigitFromLast(n, k)
{
// If k is less than equal to 0
if (k <= 0)
{
document.write(-1);
return;
}
// Convert integer into string
var temp = String(n);
// If k is greater than length of the
// string temp
if (k > temp.length)
{
document.write(-1);
}
// Print the k digit from last
else
{
var req = temp.charAt(temp.length - k)
// Convert to number again
document.write(Number(req));
}
}
// Driver code
// Given Input
var n = 2354;
var k = 2;
// Function call
kthDigitFromLast(n, k);
// This code is contributed by Potta Lokesh
</script>
Output
5
时间复杂度:* O(d)，其中 d 为数字 N 中的位数。 辅助空间:* O(d)
高效方法:这个问题可以通过迭代数字 N 的位数来解决。按照以下步骤解决此问题:
- 如果 K 小于等于 0，则打印 -1 返回。
- 迭代，同时 N 和 K-1 大于 0 :
- 将 N 更新为 N/10 ，将 K 递减 1 。
- 如果 N 为 0 ，则打印 -1 ，否则打印 N%10 。
下面是上述方法的实现:
C++
// c++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the kth digit
// from last in an integer n
void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0) {
cout << -1 << endl;
return;
}
// Divide the number n by 10
// upto k-1 times
while ((k - 1) > 0 && n > 0) {
n = n / 10;
k--;
}
// If the number n is equal 0
if (n == 0) {
cout << -1 << endl;
}
// Print the right most digit
else {
cout << n % 10 << endl;
}
}
// Driver code
int main()
{
// Given Input
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
class GFG{
// Function to find the kth digit
// from last in an integer n
public static void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0)
{
System.out.println(-1);
return;
}
// Divide the number n by 10
// upto k-1 times
while ((k - 1) > 0 && n > 0)
{
n = n / 10;
k--;
}
// If the number n is equal 0
if (n == 0)
{
System.out.println(-1);
}
// Print the right most digit
else
{
System.out.println(n % 10);
}
}
// Driver code
public static void main(String[] args)
{
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
}
// This code is contributed by Potta Lokesh
Python 3
# Python program for the above approach
# Function to find the kth digit
# from last in an eger n
def kthDigitFromLast( n, k):
# If k is less than equal to 0
if (k <= 0):
print("-1")
return
# Divide the number n by 10
# upto k-1 times
while ((k - 1) > 0 and n > 0):
n = n / 10
k -= 1
# If the number n is equal 0
if (n == 0):
print("-1")
# Pr the right most digit
else:
print(int(n % 10 ))
# Driver code
if __name__ == '__main__':
# Given Input
n = 2354
k = 2
# Function call
kthDigitFromLast(n, k)
# this code is contributed by shivanisinghss2110
C
// C# program for the above approach
using System;
class GFG
{
// Function to find the kth digit
// from last in an integer n
public static void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0)
{
Console.Write(-1);
return;
}
// Divide the number n by 10
// upto k-1 times
while ((k - 1) > 0 && n > 0)
{
n = n / 10;
k--;
}
// If the number n is equal 0
if (n == 0)
{
Console.Write(-1);
}
// Print the right most digit
else
{
Console.Write(n % 10);
}
}
// Driver code
public static void Main(String[] args)
{
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
}
// This code is contributed by shivanisinghss2110
java 描述语言
<script>
// JavaScript program for the above approach
// Function to find the kth digit
// from last in an integer n
function kthDigitFromLast(n, k)
{
// If k is less than equal to 0
if (k <= 0)
{
document.write(-1);
return;
}
// Divide the number n by 10
// upto k-1 times
while ((k - 1) > 0 && n > 0)
{
n = n / 10;
k--;
}
// If the number n is equal 0
if (n == 0)
{
document.write(-1);
}
// Print the right most digit
else
{
document.write(parseInt(n % 10));
}
}
// Driver code
// Given Input
var n = 2354;
var k = 2;
// Function call
kthDigitFromLast(n, k);
// This code is contributed by Potta Lokesh
</script>
Output
5
时间复杂度:* O(d)，其中 d 为数字 N 中的位数。 辅助空间:* O(1)
高效途径:这个问题可以使用幂函数解决。按照以下步骤解决此问题:
- 如果 K 小于等于 0，则打印 -1 返回。
- 初始化一个变量，比如说除数为幂 (10，K-1)。
- 如果除数大于 N，则打印 -1，否则打印(N/除数)%10。
下面是上述方法的实现:
C++
// c++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the kth digit
// from last in an integer n
void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0) {
cout << -1 << endl;
return;
}
// Calculate kth digit using power function
int diviser = (int)pow(10, k - 1);
// If diviser is greater than n
if (diviser > n) {
cout << -1 << endl;
}
// Otherwise print kth digit from last
else {
cout << (n / diviser) % 10 << endl;
}
}
// Driver code
int main()
{
// Given Input
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
class GFG{
// Function to find the kth digit
// from last in an integer n
public static void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0)
{
System.out.println(-1);
return;
}
// Calculate kth digit using power function
int diviser = (int)Math.pow(10, k - 1);
// If diviser is greater than n
if (diviser > n)
{
System.out.println(-1);
}
// Otherwise print kth digit from last
else
{
System.out.println((n / diviser) % 10);
}
}
// Driver code
public static void main(String[] args)
{
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
}
// This code is contributed by Potta Lokesh
Python 3
# Python program for the above approach
# Function to find the kth digit
# from last in an integer n
def kthDigitFromLast(n, k):
# If k is less than equal to 0
if (k <= 0):
print("-1")
return
# Calculate kth digit using power function
diviser = int(pow(10, k - 1))
# If diviser is greater than n
if (diviser > n):
print("-1")
# Otherwise print kth digit from last
else:
print(int((n / diviser) % 10))
# Given Input
n = 2354;
k = 2;
# Function call
kthDigitFromLast(n, k);
# This code is contributed by SoumikMondal
C
// C# program for the above approach
using System;
class GFG{
// Function to find the kth digit
// from last in an integer n
public static void kthDigitFromLast(int n, int k)
{
// If k is less than equal to 0
if (k <= 0)
{
Console.Write(-1);
return;
}
// Calculate kth digit using power function
int diviser = (int)Math.Pow(10, k - 1);
// If diviser is greater than n
if (diviser > n)
{
Console.Write(-1);
}
// Otherwise print kth digit from last
else
{
Console.Write((n / diviser) % 10);
}
}
// Driver code
public static void Main(String[] args)
{
int n = 2354;
int k = 2;
// Function call
kthDigitFromLast(n, k);
}
}
// This code is contributed by shivanisinghss2110
java 描述语言
<script>
// JavaScript program for the above approach
function kthDigitFromLast(n, k)
{
// If k is less than equal to 0
if (k <= 0)
{
document.write(-1);
return;
}
// Calculate kth digit using power function
var diviser = parseInt(Math.pow(10, k - 1));
// If diviser is greater than n
if (diviser > n)
{
document.write(-1);
}
// Otherwise print kth digit from last
else
{
document.write(parseInt((n / diviser) % 10));
}
}
// Driver code
// Given Input
var n = 2354;
var k = 2;
// Function call
kthDigitFromLast(n, k);
// This code is contributed by Potta Lokesh
</script>
Output
5
时间复杂度:* O(log(K))，其中 d 为数字 N 中的位数。这个时间复杂度是由于使用幂函数计算 10 的幂。 辅助空间: O(1)*
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