Given a m x 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?

这道题目虽然是中等难度，但是其实并不难，就是col和row很烦一定要搞清楚了，最好找个纸写下来

思路，主要针对O(1) Space：

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;
            }
        }
        
    }
}