Description
A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix’s end. For example, the matrix diagonal starting from mat[2][0], where mat is a 6 x 3 matrix, includes cells mat[2][0], mat[3][1], and mat[4][2].
Given an m x n matrix mat of integers, sort each matrix diagonal in ascending order and return the resulting matrix.
Example 1:
Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]
Example 2:
Input: mat = [[11,25,66,1,69,7],[23,55,17,45,15,52],[75,31,36,44,58,8],[22,27,33,25,68,4],[84,28,14,11,5,50]]
Output: [[5,17,4,1,52,7],[11,11,25,45,8,69],[14,23,25,44,58,15],[22,27,31,36,50,66],[84,28,75,33,55,68]]
Constraints:
m == mat.length
n == mat[i].length
1 <= m, n <= 100
1 <= mat[i][j] <= 100
Solution
For the diagonal, if the nodes are in the same diagonal, they have the same x-y
. Iterate all the numbers in the same diagonal, and sort, and fill them back.
Time complexity:
o
(
m
∗
n
∗
log
(
m
∗
n
)
)
o(m*n*log (m*n))
o(m∗n∗log(m∗n))
Space complexity:
o
(
m
∗
n
)
o(m*n)
o(m∗n)
Code
class Solution:
def diagonalSort(self, mat: List[List[int]]) -> List[List[int]]:
diag_numbers = {}
m, n = len(mat), len(mat[0])
# get diag numbers
for x in range(m):
for y in range(n):
diag_key = x - y
if diag_key not in diag_numbers:
diag_numbers[diag_key] = []
diag_numbers[diag_key].append(mat[x][y])
# sort
for k, v in diag_numbers.items():
diag_numbers[k] = sorted(v, reverse=True)
# fill back
for x in range(m):
for y in range(n):
mat[x][y] = diag_numbers[x - y].pop()
return mat
原文地址:https://blog.csdn.net/sinat_41679123/article/details/136035200
本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。
如若转载,请注明出处:http://www.7code.cn/show_67205.html
如若内容造成侵权/违法违规/事实不符,请联系代码007邮箱:suwngjj01@126.com进行投诉反馈,一经查实,立即删除!