Set Matrix Zeroes

Description:

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

#include <iostream>#include <algorithm>using namespace std;//time-complexity is O(m*n), space-complexity is O(m+n)class Solution_1{public:    void setMatrixZeros(vector<vector<int> > &matrix)    {        const size_t m = matrix.size();        const size_t n = matrix[0].size();        vector<bool> row(m,false);    //标记每一行        vector<bool> col(n,false);    //标记每一列        for (size_t i=0; i < m; ++i)        {            for (size_t j=0; j < n; ++j)            {                if (matrix[i][j] == 0)                {                    row[i] = true;                    col[j] = true;                }            }        }        for (size_t i=0; i < m; ++i)            if (row[i])                fill(&matrix[i][0],&matrix[i][0]+n,0);        for (size_t j=0; j < n; ++j)        {            if (col[j])            {                for (size_t i=0; i < m; ++i)                {                    matrix[i][j] = 0;                }            }        }         //结果输出        cout<<"Solution 1 is : "<<endl;        for (size_t i = 0  ; i < m ; ++i)        {            for (size_t j = 0; j < n ; ++j)            {                cout<<matrix[i][j]<<" ";            }            cout<<endl;        }    }};//time-complexity is O(m*n), space-complexity is O(1), 借助于矩阵的第一行，第一列class Solution_2{public:    void setMatrixZeros(vector<vector<int> > &matrix)    {        const size_t m = matrix.size();        const size_t n = matrix[0].size();        //记录下第一行，第一列第一次出现0的位置(也有可能没有0)        bool row_has_zero = false;        bool col_has_zero = false;        for (size_t i = 0  ; i < m ; ++i)  //第一列        {            if (matrix[i][0] == 0)            {                col_has_zero = true;                break;            }        }        for (size_t j = 0  ; j < n ; ++j)  //第一行        {            if (matrix[0][j] == 0)              {                   row_has_zero = true;                   break;              }        }        for (size_t i = 1 ; i < m ; ++i)        {            for (size_t j = 1; j < n; ++j)            {                if (matrix[i][j] == 0)                {                    matrix[i][0] = 0;                    matrix[0][j] = 0;                }            }        }        for (size_t i = 0; i < m ; ++i)        {            for (size_t j = 0; j < n; ++j)            {                if (matrix[i][0] == 0 | matrix[0][j] == 0)                {                    matrix[i][j] = 0;                }            }        }        if (row_has_zero)            for (size_t j = 0; j < n; ++j)            {                matrix[0][j] == 0;            }        if (col_has_zero)            for (size_t i = 0;  i < m; ++i)            {                matrix[i][0] = 0;            }        //结果输出        cout<<"Solution 2 is : "<<endl;        for (size_t i = 0  ; i < m ; ++i)        {            for (size_t j = 0; j < n ; ++j)            {                cout<<matrix[i][j]<<" ";            }            cout<<endl;        }    }};int main(){    vector<vector<int> > matrix ={{1,2,0},{3,0,3},{4,7,0}};    //结果输出        cout<<"matrix is : "<<endl;        for (size_t i = 0  ; i < matrix.size() ; ++i)        {            for (size_t j = 0; j < matrix[0].size() ; ++j)            {                cout<<matrix[i][j]<<" ";            }            cout<<endl;        }    Solution_1 solution_1;    solution_1.setMatrixZeros(matrix);    Solution_2 solution_2;    solution_2.setMatrixZeros(matrix);    return 0;}