## Set Matrix Zeroes

### 描述

Given a m × n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

Follow up: Did you use extra space?

A straight forward 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?

### 分析

O(m+n)空间的方法很简单，设置两个bool数组，记录每行和每列是否存在0。

### 代码1

// Set Matrix Zeroes
// 时间复杂度O(m*n)，空间复杂度O(m+n)
class Solution {
public:
void setZeroes(vector<vector<int> > &matrix) {
const size_t m = matrix.size();
const size_t n = matrix[0].size();
vector<bool> row(m, false); // 标记该行是否存在0
vector<bool> col(n, false); // 标记该列是否存在0

for (size_t i = 0; i < m; ++i) {
for (size_t j = 0; j < n; ++j) {
if (matrix[i][j] == 0) {
row[i] = col[j] = true;
}
}
}

for (size_t i = 0; i < m; ++i) {
if (row[i])
fill(&matrix[i][0], &matrix[i][0] + n, 0);
}
for (size_t j = 0; j < n; ++j) {
if (col[j]) {
for (size_t i = 0; i < m; ++i) {
matrix[i][j] = 0;
}
}
}
}
};


### 代码2

// Set Matrix Zeroes
// 时间复杂度O(m*n)，空间复杂度O(1)
class Solution {
public:
void setZeroes(vector<vector<int> > &matrix) {
const size_t m = matrix.size();
const size_t n = matrix[0].size();
bool row_has_zero = false; // 第一行是否存在 0
bool col_has_zero = false; // 第一列是否存在 0

for (size_t i = 0; i < n; i++)
if (matrix[0][i] == 0) {
row_has_zero = true;
break;
}

for (size_t i = 0; i < m; i++)
if (matrix[i][0] == 0) {
col_has_zero = true;
break;
}

for (size_t i = 1; i < m; i++)
for (size_t j = 1; j < n; j++)
if (matrix[i][j] == 0) {
matrix[0][j] = 0;
matrix[i][0] = 0;
}
for (size_t i = 1; i < m; i++)
for (size_t j = 1; j < n; j++)
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
if (row_has_zero)
for (size_t i = 0; i < n; i++)
matrix[0][i] = 0;
if (col_has_zero)
for (size_t i = 0; i < m; i++)
matrix[i][0] = 0;
}
};