Java

/**

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

 */
public class Transpose_Matrix {

    class Solution {
        public int[][] transpose(int[][] A) {

            if (A == null || A.length == 0) {
                return A;
            }

            int m = A.length;
            int n = A[0].length;

            int[][] transPos = new int[n][m];

            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    transPos[j][i] = A[i][j];
                }
            }

            return transPos;
        }
    }
}

// OJ: https://leetcode.com/problems/transpose-matrix/
// Time: O(MN)
// Space: O(1) extra space
class Solution {
public:
    vector<vector<int>> transpose(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size();
        vector<vector<int>> ans(N, vector<int>(M));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                ans[j][i] = A[i][j];
            }
        }
        return ans;
    }
};

class Solution:
    def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
        return list(zip(*matrix))

############

class Solution:
    def transpose(self, A):
        """
        :type A: List[List[int]]
        :rtype: List[List[int]]
        """
        rows, cols = len(A), len(A[0])
        res = [[0] * rows for _ in range(cols)]
        for row in range(rows):
            for col in range(cols):
                res[col][row] = A[row][col]
        return res

Java

class Solution {
    public int[][] transpose(int[][] A) {
        int rows = A.length, columns = A[0].length;
        int[][] transpose = new int[columns][rows];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++)
                transpose[j][i] = A[i][j];
        }
        return transpose;
    }
}

// OJ: https://leetcode.com/problems/transpose-matrix/
// Time: O(MN)
// Space: O(1) extra space
class Solution {
public:
    vector<vector<int>> transpose(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size();
        vector<vector<int>> ans(N, vector<int>(M));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                ans[j][i] = A[i][j];
            }
        }
        return ans;
    }
};

class Solution:
    def transpose(self, matrix: List[List[int]]) -> List[List[int]]:
        return list(zip(*matrix))

############

class Solution:
    def transpose(self, A):
        """
        :type A: List[List[int]]
        :rtype: List[List[int]]
        """
        rows, cols = len(A), len(A[0])
        res = [[0] * rows for _ in range(cols)]
        for row in range(rows):
            for col in range(cols):
                res[col][row] = A[row][col]
        return res

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