# LeetCode – 73. 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.

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?

1. 检查0行0列是否有0
2. 检查所有行和列，如果有0，设置第1行/第1列为0
3. 分别检查第0行和第0列，如果有0，设置第1排/第1列开始为0
4. 使用之前检查的0行0列，如果有0，设置行/列为0
```public class Solution {
public void setZeroes(int[][] matrix) {
int row = matrix.length;
int col = matrix[0].length;
boolean brow = false;
boolean bcol = false;
//check if frist row
for(int i=0;i<row;i++){
if(matrix[i][0] == 0){
bcol = true;
}
}
for(int i=0;i<col;i++){
if(matrix[0][i] == 0 ){
brow = true;
}
}

//if has 0 set first item to 0
for(int i=0;i<row;i++){
for(int j=0;j<col;j++){
if(matrix[i][j] == 0){
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
//for testing, printout matrix
// for (int i = 0; i < matrix.length; i++) {
//     for (int j = 0; j < matrix[i].length; j++) {
//         System.out.print(matrix[i][j] + " ");
//     }
//     System.out.println();
// }
//change to 0 from row and col 1
for(int i=1;i<col;i++){
if(matrix[0][i] == 0){
for(int j=1;j<row;j++){
matrix[j][i] =0;
}
}
}
for(int i=1;i<row;i++){
if(matrix[i][0] == 0){
for(int j=1;j<col;j++){
matrix[i][j] =0;
}
}
}
//change row 0 and col 0
System.out.println(col);
if(bcol){
for(int j=0;j<row;j++){
matrix[j][0] =0;
}
}
if(brow){
for(int i=0;i<col;i++){
matrix[0][i] = 0;
}
}

}
}```

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