# LeetCode: Transpose Matrix C#

https://leetcode.com/problems/transpose-matrix/

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

using System;
using System.Linq;
using Microsoft.VisualStudio.TestTools.UnitTesting;

namespace ArrayQuestions
{
[TestClass]
public class TransposeMatrix
{

/// <summary>
/// 1 2 3       1 4 7
/// 4 5 6       2 5 8
/// 7 8 9       3 6 9
/// </summary>
[TestMethod]
public void Transpose3X3()
{
int[][] matrix = { new[] { 1, 2, 3 }, new[] { 4, 5, 6 }, new[] { 7, 8, 9 } };
int[][] expected = { new[] { 1, 4, 7 }, new[] { 2, 5, 8 }, new[] { 3, 6, 9 } };
int[][] output = Transpose(matrix);
for (int i = 0; i < output.GetLength(0); i++)
{
CollectionAssert.AreEqual(expected[i], output[i]);
}
}

[TestMethod]
public void Transpose2X3()
{
int[][] matrix = { new[] { 1, 2, 3 }, new[] { 4, 5, 6 } };
int[][] expected = { new[] { 1, 4 }, new[] { 2, 5 }, new[] { 3, 6} };
int[][] output = Transpose(matrix);
for (int i = 0; i < output.GetLength(0); i++)
{
CollectionAssert.AreEqual(expected[i], output[i]);
}
}
public int[][] Transpose(int[][] A)
{
int h = A.GetLength(0);
int w = A[0].GetLength(0);

//we need to allocate the matrix as transposed already
//for example 2x3 should allocate 3x2
int[][] res = new int[w][];
for (int i = 0; i < w; i++)
{
res[i] = new int[h];
}

for (int r = 0; r < h; r++)
{
for (int c = 0; c < w; c++)
{
res[c][r] = A[r][c];
}
}
return res;
}
}
}


for (int i = 0; i < w; i++)