出现在两个给定数字中的公共数字的数量
给定两个正数 N 和 M ，任务是计算存在于 N 和 M 中的位数。
示例:
输入: N = 748294，M = 34298156 输出: 4 说明:两个数字中出现的数字是{4，8，2，9}。因此，所需的计数是 4。
输入: N = 111222，M = 333444 输出: 0 说明:两个给定的数字中没有共同的数字。
方法:给定的问题可以使用 散列 来解决。按照以下步骤解决问题:
- 初始化一个变量，比如说将计数为 0 ，以存储两个数字中共同的位数。
- 初始化两个数组，比如 freq1[10] 和 freq2[10] 为 {0} ，分别存储整数 N 和 M 中的位数。
- 迭代整数 N 的位数，并将freq 1【】中的每个位数增加 1 。
- 迭代整数 M 的位数，并将freq 2【】中的每个位数增加 1 。
- 如果频率 q1[i] 和频率 q2[i] 都超过 0 ，则迭代范围【0，9】并将计数增加 1 。
- 最后，完成上述步骤后，打印获得的计数作为所需答案。
下面是上述方法的实现:
C++14
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count number of digits
// that are common in both N and M
int CommonDigits(int N, int M)
{
// Stores the count of common digits
int count = 0;
// Stores the count of digits of N
int freq1[10] = { 0 };
// Stores the count of digits of M
int freq2[10] = { 0 };
// Iterate over the digits of N
while (N > 0) {
// Increment the count of
// last digit of N
freq1[N % 10]++;
// Update N
N = N / 10;
}
// Iterate over the digits of M
while (M > 0) {
// Increment the count of
// last digit of M
freq2[M % 10]++;
// Update M
M = M / 10;
}
// Iterate over the range [0, 9]
for (int i = 0; i < 10; i++) {
// If freq1[i] and freq2[i] both exceeds 0
if (freq1[i] > 0 & freq2[i] > 0) {
// Increment count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
int main()
{
// Input
int N = 748294;
int M = 34298156;
cout << CommonDigits(N, M);
return 0;
}
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.util.*;
class GFG{
// Function to count number of digits
// that are common in both N and M
static int CommonDigits(int N, int M)
{
// Stores the count of common digits
int count = 0;
// Stores the count of digits of N
int freq1[] = new int[10];
// Stores the count of digits of M
int freq2[] = new int[10];
// Iterate over the digits of N
while (N > 0)
{
// Increment the count of
// last digit of N
freq1[N % 10]++;
// Update N
N = N / 10;
}
// Iterate over the digits of M
while (M > 0)
{
// Increment the count of
// last digit of M
freq2[M % 10]++;
// Update M
M = M / 10;
}
// Iterate over the range [0, 9]
for(int i = 0; i < 10; i++)
{
// If freq1[i] and freq2[i] both exceeds 0
if (freq1[i] > 0 & freq2[i] > 0)
{
// Increment count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
public static void main(String[] args)
{
// Input
int N = 748294;
int M = 34298156;
System.out.print(CommonDigits(N, M));
}
}
// This code is contributed by gauravrajput1
Python 3
# Python3 program for the above approach
# Function to count number of digits
# that are common in both N and M
def CommonDigits(N, M):
# Stores the count of common digits
count = 0
# Stores the count of digits of N
freq1 = [0] * 10
# Stores the count of digits of M
freq2 = [0] * 10
# Iterate over the digits of N
while (N > 0):
# Increment the count of
# last digit of N
freq1[N % 10] += 1
# Update N
N = N // 10
# Iterate over the digits of M
while (M > 0):
# Increment the count of
# last digit of M
freq2[M % 10] += 1
# Update M
M = M // 10
# Iterate over the range [0, 9]
for i in range(10):
# If freq1[i] and freq2[i] both exceeds 0
if (freq1[i] > 0 and freq2[i] > 0):
# Increment count by 1
count += 1
# Return the count
return count
# Driver Code
if __name__ == '__main__':
# Input
N = 748294
M = 34298156
print (CommonDigits(N, M))
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
class GFG{
// Function to count number of digits
// that are common in both N and M
static int CommonDigits(int N, int M)
{
// Stores the count of common digits
int count = 0;
// Stores the count of digits of N
int[] freq1 = new int[10];
// Stores the count of digits of M
int[] freq2 = new int[10];
// Iterate over the digits of N
while (N > 0)
{
// Increment the count of
// last digit of N
freq1[N % 10]++;
// Update N
N = N / 10;
}
// Iterate over the digits of M
while (M > 0)
{
// Increment the count of
// last digit of M
freq2[M % 10]++;
// Update M
M = M / 10;
}
// Iterate over the range [0, 9]
for(int i = 0; i < 10; i++)
{
// If freq1[i] and freq2[i]
// both exceeds 0
if (freq1[i] > 0 & freq2[i] > 0)
{
// Increment count by 1
count++;
}
}
// Return the count
return count;
}
// Driver code
static void Main()
{
// Input
int N = 748294;
int M = 34298156;
Console.WriteLine(CommonDigits(N, M));
}
}
// This code is contributed by sanjoy_62
java 描述语言
<script>
// javascript program for the above approach
// Function to count number of digits
// that are common in both N and M
function CommonDigits(N,M)
{
// Stores the count of common digits
var count = 0;
// Stores the count of digits of N
var freq1 = Array(10).fill(0);
// Stores the count of digits of M
var freq2 = Array(10).fill(0);
// Iterate over the digits of N
while (N > 0) {
// Increment the count of
// last digit of N
freq1[N % 10]++;
// Update N
N = Math.floor(N / 10);
}
// Iterate over the digits of M
while (M > 0) {
// Increment the count of
// last digit of M
freq2[M % 10]++;
// Update M
M = Math.floor(M / 10);
}
var i;
// Iterate over the range [0, 9]
for (i = 0; i < 10; i++) {
// If freq1[i] and freq2[i] both exceeds 0
if (freq1[i] > 0 & freq2[i] > 0) {
// Increment count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
// Input
var N = 748294;
var M = 34298156;
document.write(CommonDigits(N, M));
</script>
Output:
4
时间复杂度: O(位数(N)+位数(M)) 辅助空间: O(10)
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