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

矩阵置空

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(m n) 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?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(m n) 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?

我们考虑就用原数组的第一行第一列来记录各行各列是否有0，
1、先扫描第一行第一列，如果有0，则将各自的flag设置为true
2、然后扫描除去第一行第一列的整个数组，如果有0，则将对应的第一行和第一列的数字赋0
3、再次遍历除去第一行第一列的整个数组，如果对应的第一行和第一列的数字有一个为0，则将当前值赋0
4、最后根据第一行第一列的flag来更新第一行第一列
代码如下：

class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) 
    {
        if (matrix.empty() || matrix[0].empty()) 
            return;
        int m = matrix.size();
        int n = matrix[0].size();

        bool rowZero = false, colZero = false;
        for (int i = 0; i < m; ++i) 
        {
            if (matrix[i][0] == 0) 
                colZero = true;
        }
        for (int i = 0; i < n; ++i) 
        {
            if (matrix[0][i] == 0) 
                rowZero = true;
        } 
        for (int i = 1; i < m; ++i)
        {
            for (int j = 1; j < n; ++j)
            {
                if (matrix[i][j] == 0) 
                {
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
            }
        }
        for (int i = 1; i < m; ++i) 
        {
            for (int j = 1; j < n; ++j)
            {
                if (matrix[0][j] == 0 || matrix[i][0] == 0)
                {
                    matrix[i][j] = 0;
                }
            }
        }
        if (rowZero) 
        {
            for (int i = 0; i < n; ++i)
                matrix[0][i] = 0;
        }
        if (colZero) 
        {
            for (int i = 0; i < m; ++i)
                matrix[i][0] = 0;
        }
    }
};