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 = -1
j = 0
for _ in range(m n-1):
if i != m -1:
i = 1
else:
j = 1
current_row = i
current_column = j
temp = []
while current_row > -1 and current_column < n:
temp.append(mat[current_row][current_column])
current_row -= 1
current_column = 1
if (i j) % 2 == 1:
result = temp[::-1]
else:
result = temp
return result
Reference
- https://leetcode.com/problems/diagonal-traverse/