[LeetCode Daily Question] - 1572. The sum of the diagonal elements of the matrix

One [topic category]

• matrix

Two [question difficulty]

• Simple

Three [topic number]

• 1572. The sum of matrix diagonal elements

Four [title description]

• Given a square matrix mat, please return the sum of the diagonal elements of the matrix.
• Please return the sum of the elements on the main diagonal of the matrix and the elements on the subdiagonal but not on the main diagonal.

Five [topic examples]

• Example 1:

  • 

  • Input: mat = [[1,2,3],
                         [4,5,6],
                         [7,8,9]]

  • Output: 25

  • Explanation: The sum of the diagonals is: 1 + 5 + 9 + 3 + 7 = 25. Note that the element mat[1][1] = 5 will only be evaluated once.

• Example 2:

  • Input: mat = [[1,1,1,1],
                          [1,1,1,1],
                          [1,1,1,1],
                          [1,1,1,1]]
  • Output: 8

• Example 3:

  • Input: mat = [[5]]
  • Output: 5

Six [topic prompt]

• $n==mat.length==mat[i].length$
• $1<=n<=100$
• $1<=mat[i][j]<=100$

Seven [problem-solving ideas]

• Define i to traverse the number of rows of the two-dimensional array, and j to traverse the number of columns of the two-dimensional array
• If i==j, it means the element of the main diagonal
• If i+j==n-1, it means that it is an element of the sub-diagonal
• Use ||to judge, so that the elements of the main diagonal and the sub-diagonal will not be added once more, because the array is only traversed once
• Then sum them, and finally return the result

Eight 【Time Frequency】

• Time complexity: $O(n^2)$，$n$ is the length of the incoming square array
• Space complexity: $O(1)$

Nine [code implementation]

1. Java language version

class Solution {
    
    
    public int diagonalSum(int[][] mat) {
    
    
        int n = mat.length;
        int res = 0;
        for(int i = 0;i < n;i++){
    
    
            for(int j = 0;j < n;j++){
    
    
                if(i == j || i + j == n - 1){
    
    
                    res += mat[i][j];
                }
            }
        }
        return res;
    }
}

2. C language version

int diagonalSum(int** mat, int matSize, int* matColSize)
{
    
    
    int n = matSize;
    int res = 0;
    for(int i = 0;i < n;i++)
    {
    
    
        for(int j = 0;j < n;j++)
        {
    
    
            if(i == j || i + j == n - 1)
            {
    
    
                res += mat[i][j];
            }
        }
    }
    return res;
}

3. Python language version

class Solution:
    def diagonalSum(self, mat: List[List[int]]) -> int:
        n = len(mat)
        res = 0
        for i in range(0,n):
            for j in range(0,n):
                if i == j or i + j == n - 1:
                    res += mat[i][j]
        return res

4. C++ language version

class Solution {
    
    
public:
    int diagonalSum(vector<vector<int>>& mat) {
    
    
        int n = mat.size();
        int res = 0;
        for(int i = 0;i < n;i++){
    
    
            for(int j = 0;j < n;j++){
    
    
                if(i == j || i + j == n - 1){
    
    
                    res += mat[i][j];
                }
            }
        }
        return res;
    }
};

Ten【Submission Results】

1. Java language version
   

2. C language version
   

3. Python language version
   

4. C++ language version