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• /**

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;
}
}
}

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

class Solution {
public int[][] transpose(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
int[][] ans = new int[n][m];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
}
}

• // 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

• func transpose(matrix [][]int) [][]int {
m, n := len(matrix), len(matrix[0])
ans := make([][]int, n)
for i := range ans {
ans[i] = make([]int, m)
for j := range ans[i] {
ans[i][j] = matrix[j][i]
}
}
return ans
}

• /**
* @param {number[][]} matrix
* @return {number[][]}
*/
var transpose = function (matrix) {
const m = matrix.length;
const n = matrix[0].length;
const ans = new Array(n).fill(0).map(() => new Array(m).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
};


• 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;
}
}

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

class Solution {
public int[][] transpose(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
int[][] ans = new int[n][m];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
}
}

• // 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

• func transpose(matrix [][]int) [][]int {
m, n := len(matrix), len(matrix[0])
ans := make([][]int, n)
for i := range ans {
ans[i] = make([]int, m)
for j := range ans[i] {
ans[i][j] = matrix[j][i]
}
}
return ans
}

• /**
* @param {number[][]} matrix
* @return {number[][]}
*/
var transpose = function (matrix) {
const m = matrix.length;
const n = matrix[0].length;
const ans = new Array(n).fill(0).map(() => new Array(m).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
};