73.set-matrix-zeroes
Statement
Metadata
- Link: 矩阵置零
- Difficulty: Medium
- Tag:
数组
哈希表
矩阵
给定一个 m x n
的矩阵,如果一个元素为 0 ,则将其所在行和列的所有元素都设为 0 。请使用 原地 算法。
示例 1:
输入:matrix = [[1,1,1],[1,0,1],[1,1,1]]
输出:[[1,0,1],[0,0,0],[1,0,1]]
示例 2:
输入:matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
输出:[[0,0,0,0],[0,4,5,0],[0,3,1,0]]
提示:
m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-231 <= matrix[i][j] <= 231 - 1
进阶:
- 一个直观的解决方案是使用
O(mn)
的额外空间,但这并不是一个好的解决方案。 - 一个简单的改进方案是使用
O(m + n)
的额外空间,但这仍然不是最好的解决方案。 - 你能想出一个仅使用常量空间的解决方案吗?
Metadata
- Link: Set Matrix Zeroes
- Difficulty: Medium
- Tag:
Array
Hash Table
Matrix
Given an m x n
integer matrix matrix
, if an element is 0
, set its entire row and column to 0
's.
You must do it in place.
Example 1:
Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[1,0,1],[0,0,0],[1,0,1]]
Example 2:
Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
Constraints:
m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-231 <= matrix[i][j] <= 231 - 1
Follow up:
- A straightforward solution using
O(mn)
space is probably a bad idea. - A simple improvement uses
O(m + n)
space, but still not the best solution. - Could you devise a constant space solution?
Solution
from typing import List
class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
ld = {}
cd = {}
for i, l in enumerate(matrix):
for j, c in enumerate(l):
if c == 0:
ld[i] = True
cd[j] = True
for i, l in enumerate(matrix):
for j, c in enumerate(l):
if (i in ld.keys() and ld[i]) or (j in cd.keys()) and cd[j]:
matrix[i][j] = 0
最后更新: October 11, 2023