766. Toeplitz Matrix*
https://leetcode.com/problems/toeplitz-matrix/
题目描述
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:
In the above grid, the diagonals are:
"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]".
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].
C++ 实现 1
class Solution {
public:
bool isToeplitzMatrix(vector<vector<int>>& matrix) {
int rows = matrix.size(), cols = matrix[0].size();
for (int i = 1; i < rows; i++) {
for (int j = 1; j < cols; ++j) {
if (matrix[i][j] != matrix[i - 1][j - 1])
return false;
}
}
return true;
}
};
C++ 实现 2
class Solution {
public:
bool isToeplitzMatrix(vector<vector<int>>& matrix) {
if (matrix.empty() || matrix[0].empty()) return true;
int m = matrix.size(), n = matrix[0].size();
// 从左下角向上
for (int i = m - 1; i >= 0; -- i) {
for (int j = 0; j < n - 1; ++ j) {
if (i + 1 < m && matrix[i][j] != matrix[i + 1][j + 1]) return false;
}
}
// 从右上角向左
for (int j = n - 1; j >= 0; -- j) {
for (int i = 0; i < m - 1; ++ i) {
if (j + 1 < n && matrix[i][j] != matrix[i + 1][j + 1]) return false;
}
}
return true;
}
};
