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(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?

大意：当[i][j]==0 当前行列都为0

要是想要常量级的空间，想到的是用二进制存储行列序号

WA了

然后，看了别人的代码才知道居然可以这么写QAQ

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        int flagi=0,flagj=0;
        int n=matrix.size();
        int m=matrix[0].size();
        for(int i=0;i<n;i++){
            if(matrix[i][0]==0)
                flagi=1;
        }
        for(int i=0;i<m;i++){
            if(matrix[0][i]==0)
                flagj=1;
        }
        for(int i=1;i<n;i++){
            for(int j=1;j<m;j++){
                if(matrix[i][j]==0)
                    matrix[0][j]=0,matrix[i][0]=0;
            }
        }
        for(int i=1;i<n;i++){
            if(matrix[i][0]==0)
                for(int j=1;j<m;j++)
                    matrix[i][j]=0;
        }
        for(int i=1;i<m;i++){
            if(matrix[0][i]==0)
                for(int j=1;j<n;j++)
                    matrix[j][i]=0;
        }
        if(flagi)
            for(int i=0;i<n;i++)
                matrix[i][0]=0;
        if(flagj)
            for(int i=0;i<m;i++)
                matrix[0][i]=0;
    }
};