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.

Have you met this question in a real interview?  

Yes

Example

Given a matrix

[
  [1,2],
  [0,3]
],

return [ [0,2], [0,0] ]

Challenge

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?

O(m + n)的空间复杂度的实现，这就不多说了；这里主要讲下如何做到常数空间，采用的方法是位图，用vector<unsigned> row, col中的每个数的每一位对应相应的行，所以空间复杂度为O(n/32 + m/32)，这样1000 * 1000的矩阵的空间也才60多个整数空间，代码如下：

class Solution {
public:
    /**
     * @param matrix: A list of lists of integers
     * @return: Void
     */
    //位向量，定义两个位向量
    void setZeroes(vector<vector<int> > &matrix) {
        // write your code here
        int n = matrix.size();
        if(n < 1)
            return;
        int m = matrix[0].size();
        int Max = max(m / 32, n / 32) + 1;
        vector<unsigned> row(Max, ~0), col(Max, ~0);
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < m; ++j){
                if(matrix[i][j] == 0){
                    row[i >> 5] = row[i >> 5] & ~(1 << (i % 32));
                    col[j >> 5] = col[j >> 5] & ~(1 << (j % 32));
                }
            }
        }
        
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < m; ++j)
                if(!((row[i >> 5] & (1 << (i % 32))) && (col[j >> 5] & (1 << (j % 32)))))
                    matrix[i][j] = 0;
        }
        
    }
};