【LeetCode】73. Set Matrix Zeroes (2 solutions)

Set Matrix Zeroes

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?

解法一：

使用数组分别记录需要置零的行列。然后根据数组信息对相应行列置零。

空间复杂度O(m+n)

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

        int m = matrix.size();
        int n = matrix[].size();

        vector<bool> row(m, false);
        vector<bool> col(n, false);

        for(int i = ; i < m; i ++)
        {
            for(int j = ; j < n; j ++)
            {
                if(matrix[i][j] == )
                {
                    row[i] = true;
                    col[j] = true;
                }
            }
        }

        for(int i = ; i < m; i ++)
        {
            for(int j = ; j < n; j ++)
            {
                if(row[i] == true)
                    matrix[i][j] = ;
                if(col[j] == true)
                    matrix[i][j] = ;
            }
        }
    }
};

解法二：

使用第一行和第一列记录该行和该列是否应该置零。

对于由此覆盖掉的原本信息，只要单独遍历第一行第一列判断是否需要置零即可。

空间复杂度O(1)

class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
        if(matrix.empty() || matrix[].empty())
            return;
        int m = matrix.size();
        int n = matrix[].size();
        bool col0 = false;
        bool row0 = false;
        for(int i = ; i < m; i ++)
        {
            if(matrix[i][] == )
            {
                col0 = true;
                break;
            }
        }
        for(int j = ; j < n; j ++)
        {
            if(matrix[][j] == )
            {
                row0 = true;
                break;
            }
        }
        for(int i = ; i < m; i ++)
        {
            for(int j = ; j < n; j ++)
            {
                if(matrix[i][j] == )
                {
                    matrix[][j] = ;
                    matrix[i][] = ;
                }
            }
        }
        for(int i = ; i < m; i ++)
        {
            if(matrix[i][] == )
            {
                for(int j = ; j < n; j ++)
                {
                    matrix[i][j] = ;
                }
            }
        }
        for(int j = ; j < n; j ++)
        {
            if(matrix[][j] == )
            {
                for(int i = ; i < m; i ++)
                {
                    matrix[i][j] = ;
                }
            }
        }
        if(col0 == true)
        {
            for(int i = ; i < m; i ++)
            {
                matrix[i][] = ;
            }
        }
        if(row0 == true)
        {
            for(int j = ; j < n; j ++)
            {
                matrix[][j] = ;
            }
        }
    }
};