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[].size(), false);
         row.resize(matrix.size(), false);
         for (int i = ; i < matrix.size(); ++i)
         {
             for (int j = ; j < matrix[i].size();++j)
             {
                 if(matrix[i][j]==)
                 {
                     col[j]=true;
                     row[i]=true;
                 }
             }
         }
         for (int i = ; i < matrix.size(); ++i)
         {
             for (int j = ; j < matrix[i].size();++j)
             {
                 if(col[j]||row[i])
                 {
                     matrix[i][j]=;
                 }
             }
         }
     }
 };

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[].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 = ; i < row; i++)
             if ( == matrix[i][]) {
                 col_flg = true;
                 break;
             }
         for (int i = ; i < col; i++)
             if ( == matrix[][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 = ; i < row; i++)
             for (int j = ; j < col; j++)
                 if ( == matrix[i][j]) {
                     matrix[i][] = ;
                     matrix[][j] = ;
                 }
         //根据记录清零
         for (int i = ; i < row; i++)
             for (int j = ; j < col; j++)
                 if ( == matrix[i][] ||  == matrix[][j])
                     matrix[i][j] = ;
         //最后处理第一行
         if (row_flg)
             for (int i = ; i < col; i++)
                 matrix[][i] = ;
         if (col_flg)
             for (int i = ; i < row; i++)
                 matrix[i][] = ;
     }
 };