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?

难度：medium

题目：给定m * n的矩阵，如果某一元素为0，则将其所在行及列都设为0。在原矩阵上执行。

一种直接的解法是空间复杂度为O(mn)一种简单的改进是空间复杂度为O(m + n),但是仍不是最好的。是否可以用常量空间给出解法。

思路：先统计首行，首列是否含有0。然后用首行首列来记录其它行列。

Runtime: 1 ms, faster than 99.98% of Java online submissions for Set Matrix Zeroes.

class Solution {    public void setZeroes(int[][] matrix) {        int m = matrix.length, n = matrix[0].length;        int row0 = 1, column0 = 1;        // first row        for (int i = 0; i < n; i++) {            if (0 == matrix[0][i]) {                row0 = 0;            }        }        // first column        for (int i = 0; i < m; i++) {            if (0 == matrix[i][0]) {                column0 = 0;            }        }        // other rows and columns        for (int i = 1; i < m; i++) {            for (int j = 1; j < n; j++) {                if (0 == matrix[i][j]) {                    matrix[i][0] = 0;                    matrix[0][j] = 0;                }            }        }        // set 0 for other rows and columns        for (int i = 1; i < m; i++) {            for (int j = 1; j < n; j++) {                if (0 == matrix[0][j] || 0 == matrix[i][0]) {                    matrix[i][j] = 0;                }            }        }        // set 0 for first row        for (int i = 0; 0 == row0 && i < n; i++) {            matrix[0][i] = 0;        }        // set 0 for first column        for (int i = 0; 0 == column0 && i < m; i++) {            matrix[i][0] = 0;        }    }}