LeetCode 867. 转置矩阵（C、C++、python）

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给定一个矩阵 A， 返回 A 的转置矩阵。

矩阵的转置是指将矩阵的主对角线翻转，交换矩阵的行索引与列索引。

示例 1：

输入：[[1,2,3],[4,5,6],[7,8,9]]
输出：[[1,4,7],[2,5,8],[3,6,9]]

示例 2：

输入：[[1,2,3],[4,5,6]]
输出：[[1,4],[2,5],[3,6]]

提示：

1 <= A.length <= 1000

1 <= A[0].length <= 1000

C语言

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/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *columnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** transpose(int** A, int ARowSize, int *AColSizes, int** columnSizes, int* returnSize) 
{
    int m=ARowSize;
    int n=*AColSizes;
    *returnSize=n;
    int** res=(int**)malloc(sizeof(int*)*n);
    *columnSizes=malloc(sizeof(int)*n);     
    for(int i=0;i<n;i++)
    {
        int* newone=(int*)malloc(sizeof(int)*m);     
        for(int j=0;j<m;j++)
        {
            newone[j]=A[j][i];
        }       
        res[i]=newone;               
        (*columnSizes)[i]=m;
    }
    return res;
}

C++

class Solution {
public:
    vector<vector<int>> transpose(vector<vector<int>>& A) 
    {
        int m=A.size();
        int n=A[0].size();
        vector<vector<int>> res(n);
        for(int i=0;i<n;i++)
        {
            res[i].resize(m);
        }            
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                res[i][j]=A[j][i];
            }
        }
        return res;
    }
};

python

class Solution:
    def transpose(self, A):
        """
        :type A: List[List[int]]
        :rtype: List[List[int]]
        """
        m=len(A)
        n=len(A[0])
        res=[[None for i in range(m)] for j in range(n)]
        for i in range(n):
            for j in range(m):
                res[i][j]=A[j][i]
        return res;