方阵的两条对角线上的逐行公共元素

给定一个方阵，找出在同一行中相同且在主对角线和次对角线中相同的数字的计数。
例子：

Input : 1 2 1
        4 5 2
        0 5 1
Output : 2
Primary diagonal is 1 5 1
Secondary diagonal is 1 5 0
Two elements (1 and 5) match 
in two diagonals and same.

Input : 1 0 0
        0 1 0
        0 0 1
Output : 1
Primary diagonal is 1 1 1
Secondary diagonal is 0 1 0
Only one element is same.

我们可以在 O(n) 时间、O(1) 空间和一次遍历中实现这一点。我们可以在主对角线的第 i 行中找到当前元素作为 mat[i][i]，在次对角线的第 i 个元素作为 mat[i][ni-1]。

C++

// CPP program to find common elements in
// two diagonals.
#include 
#define MAX 100
using namespace std;
 
// Returns count of row wise same
// elements in two diagonals of
// mat[n][n]
int countCommon(int mat[][MAX], int n)
{
    int res = 0;
    for (int i=0;i


Java
// Java program to find common
// elements in two diagonals.
import java.io.*;
 
class GFG
{
    int MAX = 100;
     
    // Returns count of row wise same elements
    // in two diagonals of mat[n][n]
    static int countCommon(int mat[][], int n)
    {
        int res = 0;
        for (int i = 0; i < n; i++)
            if (mat[i][i] == mat[i][n - i - 1])
                res++;
        return res;
    }
 
    // Driver Code
    public static void main(String args[])throws IOException
    {
        int mat[][] = {{1, 2, 3},
                       {4, 5, 6},
                       {7, 8, 9}};
        System.out.println(countCommon(mat, 3));
    }
}
 
// This code is contributed by Anshika Goyal.

Python3
# Python3 program to find common
# elements in two diagonals.
 
Max = 100
 
# Returns count of row wise same
# elements in two diagonals of
# mat[n][n]
def countCommon(mat, n):
    res = 0
     
    for i in range(n):
         
        if mat[i][i] == mat[i][n-i-1] :
            res = res + 1
    return res    
 
# Driver Code
mat = [[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]]
 
print(countCommon(mat, 3))
 
# This code is contributed by Anant Agarwal.

C#
// C# program to find common
// elements in two diagonals.
using System;
 
class GFG {
     
    // Returns count of row wise same
    // elements in two diagonals of
    // mat[n][n]
    static int countCommon(int [,]mat, int n)
    {
        int res = 0;
         
        for (int i = 0; i < n; i++)
            if (mat[i,i] == mat[i,n - i - 1])
                res++;
                 
        return res;
    }
 
    // Driver Code
    public static void Main()
    {
        int [,]mat = {{1, 2, 3},
                      {4, 5, 6},
                      {7, 8, 9}};
        Console.WriteLine(countCommon(mat, 3));
    }
}
 
// This code is contributed by vt_m.

PHP

Javascript

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