文章作者:Tyan
博客:noahsnail.com | CSDN | 简书
1. Description
2. Solution
**解析:**Version 1,根据矩阵对角线的规律,对角线的数量为m+n-1
,对角线的起点(左下到右上)为第一列加上最后一行的数据,顺序为行数一次加1
,列不变,到左下角后,行数不变,列数依次加1
,对角线上的数据依次为行减1
,列加1
。由于是对角线遍历是连续的,因此,行数加列数的和为奇数时,需要反转对角线数据。
- Version 1
class Solution:def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:m = len(mat)n = len(mat[0])result = []i = -1j = 0for _ in range(m+n-1):if i != m -1:i += 1else:j += 1current_row = icurrent_column = jtemp = []while current_row > -1 and current_column < n:temp.append(mat[current_row][current_column])current_row -= 1current_column += 1if (i + j) % 2 == 1:result += temp[::-1]else:result += tempreturn result
Reference
- https://leetcode.com/problems/diagonal-traverse/