leetcode算法：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 = 4
Output: 
[[1,2,3,4]]
Explanation:
The row-traversing of 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 = 4
Output: 
[[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. 

这道题描述的需求是：
需要我们重构一个矩阵
提供给我们某一个矩阵，然后再给我们两个整数r和c 表示希望我们重构成r 行 c 列 的矩阵
如果不能重构返回原行列式

思想就是：
首先要判断一下 如果r*c等于给定矩阵的元素数量，那就能重构
　　重构 把所有元素取出来 再按照要求生成新的矩阵就可以了

我的python代码：

 class Solution(object):
     def matrixReshape(self, nums, r, c):
         """
         :type nums: List[List[int]]
         :type r: int
         :type c: int
         :rtype: List[List[int]]
         """
         if r*c == len(nums)*len(nums[0]):
             newList = [ i for j in range(len(nums) ) for i in nums[j]  ]
             return  [ [ newList.pop(0) for i in range(c) ] for j in range(r) ]
         else:
             return nums

 if __name__ == '__main__':
     s = Solution()
     res = s.matrixReshape([[1,2],[3,4]],1,4)
     #res = s.matrixReshape([[1, 2], [3, 4]],4,1)
     print(res)