題目
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.
難度
Easy
方法
轉(zhuǎn)換后的矩陣和之前的矩陣不一樣大時楷力，返回之前的矩陣见坑。否則將之前矩陣的數(shù)據(jù)依次填入新的矩陣中
python代碼
class Solution(object):
def matrixReshape(self, nums, r, c):
"""
:type nums: List[List[int]]
:type r: int
:type c: int
:rtype: List[List[int]]
"""
numsRows = len(nums)
numsCols = len(nums[0])
if numsRows*numsCols != r*c:
return nums
result = [[0 for col in range(c)] for row in xrange(r)]
i = 0
while i < r*c:
result[i/c][i%c] = nums[i/numsCols][i%numsCols]
i += 1
return result
assert Solution().matrixReshape([[1,2],[3,4]], 1, 4) == [[1,2,3,4]]
assert Solution().matrixReshape([[1,2],[3,4]], 2, 4) == [[1,2],[3,4]]
assert Solution().matrixReshape([[1,2,3,4]], 2, 2) == [[1,2],[3,4]]
