【LeetCode】73. Set Matrix Zeroes (2 solutions)

Set Matrix Zeroes

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

click to show follow up.

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?

解法一：

使用数组分别记录需要置零的行列。然后根据数组信息对相应行列置零。

空间复杂度O(m+n)

class Solution {
public:
void setZeroes(vector<vector<int> > &matrix) {
if(matrix.empty() || matrix[0].empty())
return;

int m = matrix.size();
int n = matrix[0].size();

vector<bool> row(m, false);
vector<bool> col(n, false);

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

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

解法二：

使用第一行和第一列记录该行和该列是否应该置零。

对于由此覆盖掉的原本信息，只要单独遍历第一行第一列判断是否需要置零即可。

空间复杂度O(1)

class Solution {
public:
void setZeroes(vector<vector<int> > &matrix) {
if(matrix.empty() || matrix[0].empty())
return;
int m = matrix.size();
int n = matrix[0].size();
bool col0 = false;
bool row0 = false;
for(int i = 0; i < m; i ++)
{
if(matrix[i][0] == 0)
{
col0 = true;
break;
}
}
for(int j = 0; j < n; j ++)
{
if(matrix[0][j] == 0)
{
row0 = true;
break;
}
}
for(int i = 1; i < m; i ++)
{
for(int j = 1; j < n; j ++)
{
if(matrix[i][j] == 0)
{
matrix[0][j] = 0;
matrix[i][0] = 0;
}
}
}
for(int i = 1; i < m; i ++)
{
if(matrix[i][0] == 0)
{
for(int j = 1; j < n; j ++)
{
matrix[i][j] = 0;
}
}
}
for(int j = 1; j < n; j ++)
{
if(matrix[0][j] == 0)
{
for(int i = 1; i < m; i ++)
{
matrix[i][j] = 0;
}
}
}
if(col0 == true)
{
for(int i = 0; i < m; i ++)
{
matrix[i][0] = 0;
}
}
if(row0 == true)
{
for(int j = 0; j < n; j ++)
{
matrix[0][j] = 0;
}
}
}
};