[LeetCode] Set Matrix Zeroes

Problem:

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?

Solution1:O(mn)+ O(m+n) space

public class Solution {
    public void setZeroes(int[][] matrix) {
        if(matrix == null || matrix.length ==0 || matrix[0].length == 0)
            return;
            
        boolean row[] = new boolean[matrix.length];
        boolean column[] = new boolean[matrix[0].length];
        
        for(int i=0; i< row.length;i++){
            for(int j=0; j< column.length; j++){
                if(matrix[i][j] ==0){
                    row[i] = true;
                    column[j] = true;
                }       
            }
        }
        
        for(int i=0; i< row.length;i++){
            for(int j=0; j< column.length; j++){
                if(row[i] || column[j]){
                    matrix[i][j] =0;
                }       
            }
        }
    }
}

Solution2 :O(m*n),

public class Solution {
    public void setZeroes(int[][] matrix) {

        boolean firstRow=false;
        boolean firstColumn=false;
        for(int i=0;i<matrix.length;i++){
            if(matrix[i][0]==0){
                firstColumn = true;
                break;
            }
        }
        for(int i=0;i<matrix[0].length;i++){
            if(matrix[0][i]==0){
                firstRow = true;
                break;
            }
        }
    
        for(int i=1;i<matrix.length;i++){
            for(int j=1;j<matrix[0].length;j++){
                if(matrix[i][j]==0){
                    matrix[i][0]=0;
                    matrix[0][j]=0;
                }
            }
        }
        
        for(int i=1;i<matrix.length;i++){
            for(int j=1;j<matrix[0].length;j++){
                if(matrix[i][0]==0||matrix[0][j]==0){
                    matrix[i][j]=0;
                }
            }
        }
        
        if(firstRow){
            for(int i=0;i<matrix[0].length;i++)
                matrix[0][i]=0;
        }
        
        if(firstColumn){
            for(int i=0;i<matrix.length;i++)
                matrix[i][0]=0;
        }
    }
}

Key concept:use first row and column to hold the entire value