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For some fixed N
, an array A
is beautiful if it is a permutation of the integers 1, 2, ..., N
, such that:
For every i < j
, there is no k
with i < k < j
such that A[k] * 2 = A[i] + A[j]
.
Given N
, return any beautiful array A
. (It is guaranteed that one exists.)
Example 1:
Input: 4
Output: [2,1,4,3]
Example 2:
Input: 5
Output: [3,1,2,5,4]
Note:
1 <= N <= 1000
思路:
比赛想到的时候是分成2部分,前面一部分,后面一部分,然后尽量2者组合不会有问题,这样的话就可以分治到2边,偶数还好办,但是奇数就卡住了。
后面看题解:不要分前半部分后半部分,直接分奇数偶数
One way is to divide into [1, N / 2] and [N / 2 + 1, N]. But it will cause problems when we merge them.
Another way is to divide into odds part and evens part. So there is no k
with A[k] * 2 = odd + even
需要注意的是:对一个符合条件的数组里面所有的数进行同一个操作,仍然是满足条件的。这也是为什么能分治的条件
class Solution:
def beautifulArray(self, N):
"""
:type N: int
:rtype: List[int]
"""
if N==1: return [1]
l=self.beautifulArray(N//2)
r=self.beautifulArray(N-N//2)
return [i*2 for i in l]+[i*2-1 for i in r]