LeetCode Weekly Contest No. 299 (1/4)

最新推荐文章于 2025-07-18 19:55:42 发布

wangye_hope

最新推荐文章于 2025-07-18 19:55:42 发布

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分类专栏： LeetCode题解文章标签： leetcode

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该博客介绍了一个编程问题，目标是检查一个给定的二维整数矩阵是否符合X-Matrix的定义，即对角线上的元素非零，其余元素为零。通过遍历矩阵并检查对角线元素和非对角线元素来实现这一判断。提供的解决方案使用Python代码实现了这一逻辑。

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Problem 1. 2319. Check if Matrix Is X-Matrix

题目描述：

A square matrix is said to be an X-Matrix if both of the following conditions hold:

All the elements in the diagonals of the matrix are non-zero.
All other elements are 0.

Given a 2D integer array grid of size n x n representing a square matrix, return true if grid is an X-Matrix. Otherwise, return false.

Example 1:

Input: grid = [[2,0,0,1],[0,3,1,0],[0,5,2,0],[4,0,0,2]]
Output: true
Explanation: Refer to the diagram above. 
An X-Matrix should have the green elements (diagonals) be non-zero and the red elements be 0.
Thus, grid is an X-Matrix.

Example 2:

Input: grid = [[5,7,0],[0,3,1],[0,5,0]]
Output: false
Explanation: Refer to the diagram above.
An X-Matrix should have the green elements (diagonals) be non-zero and the red elements be 0.
Thus, grid is not an X-Matrix.

Constraints:

n == grid.length == grid[i].length
3 <= n <= 100
0 <= grid[i][j] <= 10^5

解题思路：

这个题要判断一个矩阵是否满足下面两个条件：

1. 对角线上全不为0

2. 非对角线上全为0

我们只需要搞清楚对角线的特征。

假设(i,j)是第i行、第j列元素，则它在对角线上的充要条件为：

i+j==n-1 || i==j

搞清楚这一点，于是：

class Solution:
    def checkXMatrix(self, grid: List[List[int]]) -> bool:
        n = len(grid)
        for i, j in itertools.product(range(n), range(n)):
            if i == j or i + j == n - 1:
                if grid[i][j] == 0: return False
            else:
                if grid[i][j] != 0: return False
        return True