## Description

https://leetcode.com/problems/reshape-the-matrix/

In MATLAB, there is a very useful function called ‘reshape’, which can reshape a matrix into a new one with different size but keep its original data.

You’re given a matrix represented by a two-dimensional array, and two **positive** integers **r** and **c** representing the **row** number and **column** number of the wanted reshaped matrix, respectively.

The reshaped matrix need to be filled with all the elements of the original matrix in the same **row-traversing** order as they were.

If the ‘reshape’ operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.

**Example 1:**

Input:nums = [[1,2], [3,4]] r = 1, c = 4Output:[[1,2,3,4]]Explanation:

Therow-traversingof nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.

**Example 2:**

Input:nums = [[1,2], [3,4]] r = 2, c = 4Output:[[1,2], [3,4]]Explanation:

There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.

**Note:**

- The height and width of the given matrix is in range [1, 100].
- The given r and c are all positive.

## Explanation

Get all the items in original matrix. If the number of items is sufficient to fill in the new shape matrix, just base on new shape to add items one by one. If not, returns the original matrix.

## Python Solution

```
class Solution:
def matrixReshape(self, nums: List[List[int]], r: int, c: int) -> List[List[int]]:
numbers = []
for i in range(len(nums)):
for j in range(len(nums[0])):
numbers.append(nums[i][j])
if r * c > len(numbers):
return nums
results = []
index = 0
for i in range(r):
results.append([])
for j in range(c):
results[i].append(numbers[index])
index += 1
return results
```

- Time Complexity: O(MN).
- Space Complexity: O(MN).