766. Toeplitz Matrix斜对角矩阵
[抄题]:
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.
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
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
nums.len是行数,nums[0].len是列数
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
仔细看还是能看出规律的:
[一刷]:
数组中有index+1的情况就应该注意一下上界的范围-1了
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O(n) Space complexity: O(1)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { public boolean isToeplitzMatrix(int[][] matrix) { //cc if (matrix == null || matrix[0] == null) { return false; } //for loop for (int i = 0; i < matrix.length - 1; i++) { for (int j = 0; j < matrix[i].length - 1; j++) { if (matrix[i + 1][j + 1] != matrix[i][j]) { return false; } } } //return return true; } }