# \[73]\[中等] 矩阵置零

## 题目描述

[73. 矩阵置零](https://leetcode-cn.com/problems/set-matrix-zeroes/)

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: 
[
  [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: 
[
  [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]
]
```

进阶:

* 一个直接的解决方案是使用  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。
* 你能想出一个常数空间的解决方案吗？

## 解题思路

题解参考: [矩阵置零](https://leetcode-cn.com/problems/set-matrix-zeroes/solution/ju-zhen-zhi-ling-by-leetcode-solution-9ll7/).

对于使用$$O(1)$$空间的解法, 关键思想是使用第一行和第一列记录该行该列是否有0. 但是对于第一行, 和第一列要设置一个标志位, 为了防止自己这一行(一列)也有0的情况.

```python
class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        n, m = len(matrix), len(matrix[0])
        col0 = False

        for i in range(n):
            if matrix[i][0] == 0:
                col0 = True
            for j in range(1, m):
                if matrix[i][j] == 0:
                    matrix[0][j] = 0
                    matrix[i][0] = 0

        for i in range(n - 1, -1, -1):
            for j in range(1, m):
                if matrix[0][j] == 0 or matrix[i][0] == 0:
                    matrix[i][j] = 0
            if col0:
                matrix[i][0] = 0
```


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