N 和 M 中相同且出现在相同索引上的数字计数

给定两个数字N和M ，任务是找到N和M中相同且出现在相同索引上的数字的计数。

例子：

Input: N = 123, M = 321
Output: 1
Explanation: Digit 2 satisfies the condition

Input: N = 123, M = 111
Output: 1

方法：可以使用两指针方法解决问题。

• 转换 N 和 M 以便于遍历
• 创建两个指针，其中一个指针最初指向 N 的第一个数字，另一个指向 M 的第一个数字。
• 现在从左到右遍历两个数字并检查两个指针上的数字是否相同。
• 如果是，请增加计数。
• 每次迭代将指针加 1。
• 最后返回最终计数。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
int countDigits(int N, int M)
{
 
    // Convert N and M to string
    // for ease of traversal
    string a = to_string(N), b = to_string(M);
 
    int same_dig_cnt = 0;
 
    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.length() && b.length(); i++)
        if (a[i] == b[i])
            same_dig_cnt++;
 
    return same_dig_cnt;
}
 
// Driver code
int main()
{
 
    int N = 123, M = 321;
 
    cout << countDigits(N, M);
    return 0;

Java

// Java implementation of the above approach
class GFG {
 
  static int countDigits(int N, int M) {
 
    // Convert N and M to string
    // for ease of traversal
    String a = Integer.toString(N), b = Integer.toString(M);
 
    int same_dig_cnt = 0;
 
    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.length() && i < b.length(); i++)
      if (a.charAt(i) == b.charAt(i))
        same_dig_cnt++;
 
    return same_dig_cnt;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int N = 123, M = 321;
    System.out.println(countDigits(N, M));
  }
}
 
// This code is contributed by gfgking

Python3

# Python code for the above approach
def countDigits(N, M):
 
    # Convert N and M to string
    # for ease of traversal
    a = str(N)
    b = str(M)
 
    same_dig_cnt = 0;
 
    # Find the count of digits
    # that are same and occur on same indices
    # in both N and M.
    # Store the count in variable same_dig_cnt
    i = 0
    while(i < len(a) and len(b)):
        if (a[i] == b[i]):
            same_dig_cnt += 1
        i += 1
 
    return same_dig_cnt;
 
# Driver code
N = 123
M = 321;
 
print(countDigits(N, M));
 
# This code is contributed by Saurabh Jaiswal

C#

// C# implementation of the above approach
using System;
using System.Collections;
class GFG
{
 
  static int countDigits(int N, int M)
  {
 
    // Convert N and M to string
    // for ease of traversal
    string a = N.ToString(), b = M.ToString();
 
    int same_dig_cnt = 0;
 
    // Find the count of digits
    // that are same and occur on same indices
    // in both N and M.
    // Store the count in variable same_dig_cnt
    for (int i = 0; i < a.Length && i < b.Length; i++)
      if (a[i] == b[i])
        same_dig_cnt++;
 
    return same_dig_cnt;
  }
 
  // Driver code
  public static void Main()
  {
 
    int N = 123, M = 321;
 
    Console.Write(countDigits(N, M));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

       // JavaScript code for the above approach
       function countDigits(N, M) {
 
           // Convert N and M to string
           // for ease of traversal
           let a = (N).toString(), b = (M).toString();
 
           let same_dig_cnt = 0;
 
           // Find the count of digits
           // that are same and occur on same indices
           // in both N and M.
           // Store the count in variable same_dig_cnt
           for (let i = 0; i < a.length && b.length; i++)
               if (a[i] == b[i])
                   same_dig_cnt++;
 
           return same_dig_cnt;
       }
 
       // Driver code
       let N = 123, M = 321;
 
       document.write(countDigits(N, M));
 
      // This code is contributed by Potta Lokesh
   

输出

1

时间复杂度： O(D)，其中 D 是 N 或 M 中的最小位数
辅助空间： O(D)，其中 D 是 N 或 M 中的最大位数