LeetCode-566：Reshape the Matrix （矩阵整型）-- easy

Question

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.

问题解析：

给定用二维数组表示的矩阵nums，定值 r 和 c , 将矩阵reshape成指定的大小，返回新的二维数组；若元素个数不匹配，则返回nums。

Answer

Solution 1：

分步解决即可。

首先将List转换成一个扁平的List，利用

flat = sum(nums, [])

来实现；

判断扁平化的List长度是否与新List元素个数匹配，不匹配则返回原List；

利用zip将扁平的List重新整型为r*c的大小；利用map返回嵌套List。

class Solution(object):
def matrixReshape(self, nums, r, c):
"""
:type nums: List[List[int]]
:type r: int
:type c: int
:rtype: List[List[int]]
"""
nums_reshape = []
flat = sum(nums, [])
if len(flat) != r*c:
return nums
tuples = zip(* ([iter(flat)]*c)) # tuples： [(1,2,3),(4,5,6),(7,8,9)]
return map(list, tuples)

Solution 2：

使用itertools。

itertools.chain(*iterables)

将iterables其中的元素依次全部返回。

def matrixReshape(self, nums, r, c):
if r * c != len(nums) * len(nums[0]):
return nums
it = itertools.chain(*nums)
return [list(itertools.islice(it, c)) for _ in xrange(r)]