LeetCode-73-Set Matrix Zeroes

算法描述：

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?

解题思路：将矩阵的首行和首列用于标记该行或该列中有0元素。而首行和首列则需要单独的标记位标记该行或者列中有0元素存在。

    void setZeroes(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        bool row = false;
        bool col = false;
        for(int i=0; i < m; i++) if(matrix[i][0]==0) row = true;
        for(int j=0; j < n; j++) if(matrix[0][j]==0) col = true;
        for(int i=1; i < m; i++){
            for(int j=1; j < n; j++){
                if(matrix[i][j]==0){
                    matrix[i][0]=0;
                    matrix[0][j]=0;
                    }
            }
        }
        for(int i = 1; i < m; i++){
            for(int j =1; j < n; j++){
                if(matrix[i][0]==0 || matrix[0][j]==0){
                    matrix[i][j]=0;
                }
            }
        }
        if(row)
            for(int i=0; i < m; i++) matrix[i][0]=0;
        if(col)
            for(int i=0; i < n; i++) matrix[0][i]=0;
    }