【Leetcode】Set matrix zeros

【题目】

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

Followup:

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 thebest solution.
Could you devise a constant space solution?

【思路】

My idea is simple: store states of each row in the first of that row, and store states of each column in the first of that column. Because the state of row0 and the state of column0 would occupy the same cell, I let it be the state of row0, and use another variable "col0" for column0. In the first phase, use matrix elements to set states in a top-down way. In the second phase, use states to set matrix elements in a bottom-up way.

【代码】

void setZeroes(vector<vector<int> > &matrix) {    int col0 = 1, rows = matrix.size(), cols = matrix[0].size();    for (int i = 0; i < rows; i++) {        if (matrix[i][0] == 0) col0 = 0;        for (int j = 1; j < cols; j++)            if (matrix[i][j] == 0)                matrix[i][0] = matrix[0][j] = 0;    }    for (int i = rows - 1; i >= 0; i--) {        for (int j = cols - 1; j >= 1; j--)            if (matrix[i][0] == 0 || matrix[0][j] == 0)                matrix[i][j] = 0;        if (col0 == 0) matrix[i][0] = 0;    }}

why must bottom up?!!!!!