LeetCode刷题——Array篇（对角线相等）_在给定一个二维数组的情况下,保证所有对角线元素相等-优快云博客

本文介绍了一种简单的方法来判断一个给定的二维矩阵是否为Toeplitz矩阵。Toeplitz矩阵的特点是从左上到右下的每一条对角线上的元素都相同。文章通过示例详细解释了这一概念，并提供了一个C++实现方案，该方案只需遍历矩阵右下方的元素即可高效地完成判断。

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A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same element.

Now given an M x N matrix, return True if and only if the matrix is Toeplitz.

Example 1:

Input: matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]]
Output: True
Explanation:
1234
5123
9512

In the above grid, the diagonals are “[9]”, “[5, 5]”, “[1, 1, 1]”, “[2, 2, 2]”, “[3, 3]”, “[4]”, and in each diagonal all elements are the same, so the answer is True.
Example 2:

Input: matrix = [[1,2],[2,2]]
Output: False
Explanation:
The diagonal “[1, 2]” has different elements.
Note:

matrix will be a 2D array of integers.
matrix will have a number of rows and columns in range [1, 20].
matrix[i][j] will be integers in range [0, 99].

第一次做，针对每个对角线进行判断，看起来很麻烦，实际上只需要遍历右下角的每个元素是否和对角线的不等，不等就是不符合条件。

class Solution {
public:
    bool isToeplitzMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        for (int i = 1; i < m; i++)
            for (int j = 1; j < n; j++)
                if (matrix[i][j] != matrix[i - 1][j - 1])
                    return false;
        return true;
    }
};