# 矩阵置零
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。
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?
示例 1:
![](https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0073.Set%20Matrix%20Zeroes/images/mat1.jpg)
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:
![](https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0073.Set%20Matrix%20Zeroes/images/mat2.jpg)
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
以下错误的选项是?
## aop
### before
```cpp
#include
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
int a = 3, b = 3;
vector> matrix = vector>(a, vector(b)) = {{1, 1, 1}, {1, 0, 1}, {1, 1, 1}};
sol.setZeroes(matrix);
for (auto i : matrix)
{
for (auto j : i)
cout << j << " ";
cout << endl;
}
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
void setZeroes(vector> &matrix)
{
int row = matrix.size();
int col = matrix[0].size();
vector pos;
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
{
if (matrix[i][j] == 0)
{
pos.push_back(i);
pos.push_back(j);
}
}
}
for (int i = 1; i < pos.size(); i += 2)
{
int x = pos[i];
int y = pos[i + 1];
for (int j = 0; j < row; j++)
{
matrix[j][y] = 0;
}
for (int k = 0; k < col; k++)
{
matrix[x][k] = 0;
}
}
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
void setZeroes(vector> &matrix)
{
if (matrix.size() <= 0)
return;
int row = -1, col = -1;
int i = 0;
for (i = 0; i < matrix.size(); i++)
{
for (int j = 0; j < matrix[0].size(); j++)
{
if (matrix[i][j] == 0)
{
row = i;
col = j;
break;
}
}
if (row != -1)
break;
}
if (i == matrix.size())
return;
for (int k = i + 1; k < matrix.size(); k++)
{
for (int j = 0; j < matrix[0].size(); j++)
{
if (matrix[k][j] == 0)
{
matrix[row][j] = 0;
matrix[k][col] = 0;
}
}
}
for (i = 0; i < matrix[0].size(); i++)
{
if (i == col)
continue;
if (matrix[row][i] == 0)
for (int j = 0; j < matrix.size(); j++)
matrix[j][i] = 0;
}
for (i = 0; i < matrix.size(); i++)
{
if (i == row)
continue;
if (matrix[i][col] == 0)
for (int j = 0; j < matrix[0].size(); j++)
matrix[i][j] = 0;
}
for (i = 0; i < matrix.size(); i++)
matrix[i][col] = 0;
for (i = 0; i < matrix[0].size(); i++)
matrix[row][i] = 0;
}
};
```
### B
```cpp
class Solution
{
public:
void setZeroes(vector> &matrix)
{
bool row = false, col = false;
int m = matrix.size(), n = matrix[0].size();
for (int i = 0; i < m; ++i)
{
if (matrix[i][0] == 0)
col = true;
}
for (int i = 0; i < n; ++i)
{
if (matrix[0][i] == 0)
row = true;
}
for (int i = 1; i < m; ++i)
for (int j = 1; j < n; ++j)
{
if (matrix[i][j] == 0)
matrix[i][0] = 0, matrix[0][j] = 0;
}
for (int i = 1; i < m; ++i)
for (int j = 1; j < n; ++j)
{
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
}
if (col)
{
for (int i = 0; i < m; ++i)
matrix[i][0] = 0;
}
if (row)
{
for (int i = 0; i < n; ++i)
matrix[0][i] = 0;
}
}
};
```
### C
```cpp
class Solution
{
public:
void setZeroes(vector> &matrix)
{
int m = matrix.size(), n = matrix[0].size();
vector row(m, false), column(n, false);
int i, j;
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
if (matrix[i][j] == 0)
{
row[i] = true;
column[j] = true;
}
}
}
for (i = 0; i < m; i++)
{
if (row[i])
{
for (j = 0; j < n; j++)
matrix[i][j] = 0;
}
}
for (i = 0; i < n; i++)
{
if (column[i])
{
for (j = 0; j < m; j++)
matrix[j][i] = 0;
}
}
}
};
```