Leetcode 73 Set Matrix Zeroes

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

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Follow up:

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?

找出矩阵中的0点，并将所在行列都置0

具体要求为使用较小的空间，开始m+n的做法

class Solution {public:    void setZeroes(vector<vector<int>>& matrix) {        vector<int> x;        for(int i=0;i<matrix.size();i++)            for(int j=0;j<matrix[0].size();j++)                if(matrix[i][j]==0)                {                    x.push_back(i);                    x.push_back(j);                }        for(int i=0;i<x.size();i+=2)        {            for(int j=0;j<matrix.size();j++) matrix[j][x[i+1]]=0;            for(int j=0;j<matrix[0].size();j++) matrix[x[i]][j]=0;        }    }};

在discuss中看到的O(1)做法，将是否有0存在每一行每一列的第一个位置，

因为行列会交叉，因而会当左上角为0时会不知道到底是行还是列，

所以引入col0记录，col0为0表示是第一列产生的0

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;    }}

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