leetcode 566 Reshape the Matrix
2017-05-18 16:15
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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:
就是将矩阵按要求的行数和列数重新排列。
Answer:
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 len(nums)*len(nums[0])!=r*c:
return nums
oneArray=[]
reshaped=[[0]*c for i in range(r)]
for x in nums:
oneArray+=x
for index,item in enumerate(oneArray):
rowPlace=index/c
colPlace=index%c
reshaped[rowPlace][colPlace]=item
return reshaped
其中用到了python里的enumerate函数,也就是枚举,得到矩阵里数据的元素和下标,
具体可以看http://blog.csdn.net/zq_alen/article/details/72470371
总的思想就是先把二维矩阵变成一维矩阵,这里直接一个遍历矩阵就可以得到。
再就是通过enumerate函数将一维数组按照要求的行和列变成二维数组
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.
就是将矩阵按要求的行数和列数重新排列。
Answer:
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 len(nums)*len(nums[0])!=r*c:
return nums
oneArray=[]
reshaped=[[0]*c for i in range(r)]
for x in nums:
oneArray+=x
for index,item in enumerate(oneArray):
rowPlace=index/c
colPlace=index%c
reshaped[rowPlace][colPlace]=item
return reshaped
其中用到了python里的enumerate函数,也就是枚举,得到矩阵里数据的元素和下标,
具体可以看http://blog.csdn.net/zq_alen/article/details/72470371
总的思想就是先把二维矩阵变成一维矩阵,这里直接一个遍历矩阵就可以得到。
再就是通过enumerate函数将一维数组按照要求的行和列变成二维数组
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