LeetCode(Array&String)Set Matrix Zeroes

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 1:

Input: 
[
  [1,1,1],
  [1,0,1],
  [1,1,1]
]
Output: 
[
  [1,0,1],
  [0,0,0],
  [1,0,1]
]

Example 2:

Input: 
[
  [0,1,2,0],
  [3,4,5,2],
  [1,3,1,5]
]
Output: 
[
  [0,0,0,0],
  [0,4,5,0],
  [0,3,1,0]
]

Follow up:

• A straight forward solution using O(m**n) 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?

思路：分两部分来做，1. 0行和0列含0时，对应的标志位置为true;

2. 将不在0行和0列的为0的元素，移到0行和0列，然后将排除0行和0列之后，对剩下的元素中满足0行或者0列有0元素的行和列全部置为0。
3. 最后将对应标志位为true的0行或者0列全部置为0即可。

代码如下：

class Solution {
    public void setZeroes(int[][] matrix) {
        int row=matrix.length;
        int col=matrix[0].length;
        boolean fr=false;
        boolean fc=false;
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(matrix[i][j]==0){
                    if(i==0){
                        fr=true;
                    }
                    if(j==0){
                        fc=true;
                    }
                    matrix[i][0]=0;
                    matrix[0][j]=0;
                }
            }
        }
        for(int i=1;i<row;i++){
            for(int j=1;j<col;j++){
                if(matrix[0][j]==0 || matrix[i][0]==0){
                    matrix[i][j]=0;
                }
            }
        }
        if(fr==true){
            for(int j=0;j<col;j++){
                matrix[0][j]=0;
            }
        }
        if(fc==true){
            for(int i=0;i<row;i++){
                matrix[i][0]=0;
            }
        }
        
    }
}