set-matrix-zeroes当元素为0则设矩阵内行与列均为0

题目描述

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?

class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
        vector<bool> col,row;
        col.resize(matrix[0].size(), false);
        row.resize(matrix.size(), false);
        for (int i = 0; i < matrix.size(); ++i)
        {
            for (int j = 0; j < matrix[i].size();++j)
            {
                if(matrix[i][j]==0)
                {
                    col[j]=true;
                    row[i]=true;
                }
            }
        }
        for (int i = 0; i < matrix.size(); ++i)
        {
            for (int j = 0; j < matrix[i].size();++j)
            {
                if(col[j]||row[i])
                {
                    matrix[i][j]=0;
                }
            }
        }
    }
};

最优解法：

首先判断第一行和第一列是否有元素为0，而后利用第一行和第一列保存状态，最后根据开始的判断决定是否将第一行和第一列置0

//时间复杂度O(mn)，空间复杂度O(1)
//利用第一行和第一列的空间做记录
class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
        const int row = matrix.size();
        const int col = matrix[0].size();
        bool row_flg = false, col_flg = false;
         
        //判断第一行和第一列是否有零，防止被覆盖
        for (int i = 0; i < row; i++)
            if (0 == matrix[i][0]) {
                col_flg = true;
                break;
            }
        for (int i = 0; i < col; i++)
            if (0 == matrix[0][i]) {
                row_flg = true;
                break;
            }
        //遍历矩阵，用第一行和第一列记录0的位置
        for (int i = 1; i < row; i++)
            for (int j = 1; j < col; j++)
                if (0 == matrix[i][j]) {
                    matrix[i][0] = 0;
                    matrix[0][j] = 0;
                }
        //根据记录清零
        for (int i = 1; i < row; i++)
            for (int j = 1; j < col; j++)
                if (0 == matrix[i][0] || 0 == matrix[0][j])
                    matrix[i][j] = 0;
        //最后处理第一行
        if (row_flg)
            for (int i = 0; i < col; i++)
                matrix[0][i] = 0;
        if (col_flg)
            for (int i = 0; i < row; i++)
                matrix[i][0] = 0;
    }
};