Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

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.

提示的算法很容易就想到。constant space的算法， 自己一开始总是想找出一种操作，然后遍历一遍解决。

其实最简单的方法就是直接在matrix上记录行、列是否0的信息。这里的技巧在于，选择了在第一行和第一列上记录，用0值表示这一行（或列）为0。因为只要这一行（或列）为0，最终这个值也是要设为0的，不会影响最终的结果。如果在第一行的元素在该列没有0，那么他是否要设为0就取决于第一行是否有0. 第一列的元素同理。

1 class Solution { 2 public: 3     void setZeroes(vector
     
      
        > &matrix) { 4         int row = matrix.size(); 5         if (row == 0) return; 6         int col = matrix[0].size(); 7         if (col == 0) return; 8          9         bool firstRowZero = false, firstColZero = false;10         for (int i = 0; i < col; ++i) {11             if (matrix[0][i] == 0) {12                 firstRowZero = true;13                 break;14             }15         }16         17         for (int i = 0; i < row; ++i) {18             if (matrix[i][0] == 0) {19                 firstColZero = true;20                 break;21             }22         }23         24         for (int i = 1; i < row; ++i) {25             for (int j = 1; j < col; ++j) {26                 if (matrix[i][j] == 0) {27                     matrix[i][0] = 0;28                     matrix[0][j] = 0;29                 }30             }31         }32         33         for (int i = 1; i < row; ++i) {34             for (int j = 1; j < col; ++j) {35                 if (matrix[i][0] == 0 || matrix[0][j] == 0) {36                     matrix[i][j] = 0;37                 }38             }39         }40         41         if (firstRowZero) {42             for (int i = 0; i < col; ++i) {43                 matrix[0][i] = 0;44             }45         }46         47         if (firstColZero) {48             for (int i = 0; i < row; ++i) {49                 matrix[i][0] = 0;50             }51         }52     }53 };