LeetCode题解Transpose Matrix

1、题目描述

2、题目描述

直接申请内存,转置即可。

3、代码

 1  vector<vector<int>> transpose(vector<vector<int>>& A) {
 2         int C = A.size() ;
 3         int R = A[0].size() ;
 4         vector<vector<int>> result(R,vector<int>(C) ) ;
 5
 6         for( int i = 0 ; i < A.size() ; i++ )
 7             for( int j = 0; j < A[0].size();j++ )
 8             {
 9                 result[j][i] = A[i][j];
10             }
11         return result;
12
13     }

原文地址:https://www.cnblogs.com/wangxiaoyong/p/9291020.html

时间: 2024-11-09 03:22:54

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