LeetCode73 Set Matrix Zeroes 矩阵置零 C++

问题描述：
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?
题源：here；完整实现：here
思路：
其实这道题就是想牺牲时间来换取空间，为此我们不得不分批次的遍历输入数组。我们用第一行和第一列来存储需要置零的行和列，为了不被后面的处理覆盖原来的行列，我们需要记录第一行和第一列是否需要置零。代码如下：

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        int rowLen = matrix.size(), colLen = matrix[0].size();
        bool firstRow = false, firstCol = false;
        for (int i = 0; i < rowLen; i++){
            if (matrix[i][0] == 0){
                firstRow = true; break;
            }
        }
        for (int j = 0; j < colLen; j++){
            if (matrix[0][j] == 0){
                firstCol = true; break;
            }
        }
        for (int i = 1; i < rowLen; i++){
            for (int j = 1; j < colLen; j++){
                if (!matrix[i][j]){
                    matrix[i][0] = 0; matrix[0][j] = 0;
                }
            }
        }

        for (int i = 1; i < rowLen; i++){
            if (!matrix[i][0]){
                for (int j = 1; j < colLen; j++)
                    matrix[i][j] = 0;
            }
        }

        for (int j = 1; j < colLen; j++){
            if (!matrix[0][j]){
                for (int i = 1; i < rowLen; i++)
                    matrix[i][j] = 0;
            }
        }

        if (firstRow){
            for (int i = 0; i < rowLen; i++)
                matrix[i][0] = 0;
        }
        if (firstCol){
            for (int j = 0; j < colLen; j++)
                matrix[0][j] = 0;
        }
    }
};

虽然说这是以牺牲时间换取空间，但是似乎速度也不错：