题目描述：

Given a matrix A, return the transpose of A.

The transpose of a matrix is the matrix flipped over it's main diagonal, switching the row and column indices of the matrix.

 

Example 1:

Input: [[1,2,3],[4,5,6],[7,8,9]]Output: [[1,4,7],[2,5,8],[3,6,9]]

Example 2:

Input: [[1,2,3],[4,5,6]]Output: [[1,4],[2,5],[3,6]]

 

Note:

1. 1 <= A.length <= 1000
2. 1 <= A[0].length <= 1000

解题思路：

暴力遍历原来矩阵的每个元素，放到输出矩阵的对应位置。

由于veector的内存分配机制，由于已知输出矩阵的大小，所以在每个vector定义时指定空间大小会节省运行时间。

 

代码：

1 class Solution { 2 public: 3     vector
       
        
         > transpose(vector
         
          
           >& A) { 4         vector
           
            
              > res; 5 res.reserve(A[0].size()); 6 for (int i = 0; i < A[0].size(); ++i) { 7 vector
             
               tmp; 8 tmp.reserve(A.size()); 9 for (int j = 0; j < A.size(); ++j) {10 tmp.push_back(A[j][i]);11 }12 res.push_back(tmp);13 }14 return res;15 }16 };