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: ArrayHash TableMatrix
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