给定两个正数N和M ，任务是计算同时存在于N和M中的位数。
例子：
Input: N = 748294, M = 34298156
Output: 4
Explanation: The digits that are present in both the numbers are {4, 8, 2, 9}. Therefore, the required count is 4.
Input: N = 111222, M = 333444
Output: 0
Explanation: No common digits present in the two given numbers.
方法：可以使用Hashing解决给定的问题。请按照以下步骤解决问题：
- 初始化一个变量，例如count为0 ，以存储两个数字中的通用位数。
- 将两个数组freq1 [10]和freq2 [10]初始化为{0} ，以存储分别存在于整数N和M中的位数。
- 遍历整数N的数字，并将freq1 []中每个数字的计数增加1 。
- 遍历整数M的数字，并将freq2 []中每个数字的计数增加1 。
- 如果freq1 [i]和freq2 [i]都超过0 ，则在范围[ 0，9]上进行迭代，并将计数增加1 。
- 最后，完成上述步骤后，将获得的计数打印为所需答案。
下面是上述方法的实现：
C++14
// C++ program for the above approach
#include
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
// 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 gauravrajput1Python3
# 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 29C#
// 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输出：
4时间复杂度： O(位数(N)+位数(M))
辅助空间： O(10)