leetcode-2319-判断矩阵是否是一个 x 矩阵-爱代码爱编程
1、遍历
我们可以遍历正方形矩阵中的每一个格子,若同时满足 ( g r i d [ i ] [ j ] = = 0 ) = = ( i = = j ∣ ∣ i + j = = n − 1 ) (grid[i][j] == 0) == (i == j || i + j == n - 1) (grid[i][j]==0)==(i==j∣∣i+j==n−1)则说明不满足题意,返回false;否则返回true。
class Solution {
public:
bool checkXMatrix(vector<vector<int>> &grid) {
int n = grid.size();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if ((grid[i][j] == 0) == (i == j || i + j == n - 1)) {
return false;
}
}
}
return true;
}
};