lintcode-medium-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.

Example

Given a matrix

[
[1,2],
[0,3]
],

return
[
[0,2],
[0,0]
]

Challenge

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?

public class Solution {
/**
* @param matrix: A list of lists of integers
* @return: Void
*/
public void setZeroes(int[][] matrix) {
// write your code here

if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0)
return;

int m = matrix.length;
int n = matrix[0].length;

boolean first_col = false;
boolean first_row = false;

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

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

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

for(int i = 1; i < m; i++){
if(matrix[i][0] == 0){
for(int j = 0; j < n; j++)
matrix[i][j] = 0;
}
}

if(first_col){
for(int i = 0; i < m; i++)
matrix[i][0] = 0;
}

if(first_row){
for(int i = 0; i < n; i++)
matrix[0][i] = 0;
}

return;
}
}