【Set Matrix Zeros】cpp

题目：

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

click to show follow up.

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?

代码：

class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
            const size_t size_row = matrix.size();
            const size_t size_col = matrix[0].size();
            bool if_row0_has_zero = false;
            bool if_col0_has_zero = false;
            for (size_t col = 0; col < size_col; ++col){
                if (matrix[0][col]==0){
                    if_row0_has_zero = true;
                    break;
                }
            }
            for (size_t row = 0; row < size_row; ++row){
                if (matrix[row][0]==0){
                    if_col0_has_zero = true;
                    break;
                }
            }
            for (size_t row = 0; row < size_row; ++row){
                for (size_t col = 0; col < size_col; ++col){
                    if (matrix[row][col]==0){
                        matrix[row][0] = 0;
                        matrix[0][col] = 0;
                    }
                }
            }
            for (size_t row = 1; row < size_row; ++row){
                for (size_t col = 1; col < size_col; ++col){
                    if ( matrix[row][0]==0 || matrix[0][col]==0 ) matrix[row][col]=0;
                }
            }
            if (if_row0_has_zero) {
                for (size_t col = 0; col < size_col; ++col) matrix[0][col]=0;
            }
            if (if_col0_has_zero){
                for (size_t row = 0; row < size_row; ++row) matrix[row][0]=0;
            }
    }
};

 

Tips:

1. 算法时间复杂度上是O(n²)

2. 这里为了节省空间复杂度，用到的技巧是把第一行和第一列作为该行或该列是否含有0元素的标志位，这样就可以在常数空间内完成题目。

========================================================

第二次过这道题，思路比较清晰，代码也一次AC了。

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
            if (matrix.size()==0) return;
            // check first row
            bool first_row_zero = false;
            for ( int j=0; j<matrix[0].size(); ++j ) { 
                if (matrix[0][j]==0) {first_row_zero=true;break;}
            }
            // check frist col
            bool first_col_zero = false;
            for ( int j=0; j<matrix.size(); ++j ) { 
                if ( matrix[j][0]==0) {first_col_zero=true;break;}
            }
            // check remains
            for ( int i=1; i<matrix.size(); ++i )
            {
                for ( int j=1; j<matrix[i].size(); ++j )
                {
                    if ( matrix[i][j]==0 )
                    {
                        matrix[0][j] = 0;
                        matrix[i][0] = 0;
                    }
                }
            }
            // set row zeros
            for ( int i=1; i<matrix.size(); ++i )
            {
                if ( matrix[i][0]==0 )
                {
                    for ( int j=1; j<matrix[i].size(); ++j ) matrix[i][j] = 0;
                }
            }
            // set col zeros
            for ( int j=1; j<matrix[0].size(); ++j )
            {
                if ( matrix[0][j]==0 )
                {
                    for ( int i=1; i<matrix.size(); ++i ) matrix[i][j] = 0;
                }
            }
            if (first_row_zero)
            {
                for ( int j=0; j<matrix[0].size(); ++j ) matrix[0][j] = 0;
            }
            if ( first_col_zero)
            {
                for ( int i=0; i<matrix.size(); ++i ) matrix[i][0] = 0;
            }
    }
};

这个代码大体上没有问题，但是在一个部分是可以优化的。就是set row zeros和set col zeros两个部分可以合成一个，改一版代码如下。

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
            if (matrix.size()==0) return;
            // check first row
            bool first_row_zero = false;
            for ( int j=0; j<matrix[0].size(); ++j ) { 
                if (matrix[0][j]==0) {first_row_zero=true;break;}
            }
            // check frist col
            bool first_col_zero = false;
            for ( int j=0; j<matrix.size(); ++j ) { 
                if ( matrix[j][0]==0) {first_col_zero=true;break;}
            }
            // check remains
            for ( int i=1; i<matrix.size(); ++i )
            {
                for ( int j=1; j<matrix[i].size(); ++j )
                {
                    if ( matrix[i][j]==0 )
                    {
                        matrix[0][j] = 0;
                        matrix[i][0] = 0;
                    }
                }
            }
            // set zeros
            for ( int i=1; i<matrix.size(); ++i )
            {
                for ( int j=1; j<matrix[i].size(); ++j )
                {
                    if ( matrix[i][0]==0 || matrix[0][j]==0 ) matrix[i][j]=0;
                }
            }
            if (first_row_zero)
            {
                for ( int j=0; j<matrix[0].size(); ++j ) matrix[0][j] = 0;
            }
            if ( first_col_zero)
            {
                for ( int i=0; i<matrix.size(); ++i ) matrix[i][0] = 0;
            }
    }
};

这样改版后，代码效率提升了。