1+ """
2+ We have some permutation A of [0, 1, ..., N - 1], where N is the length of A.
3+
4+ The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j].
5+
6+ The number of local inversions is the number of i with 0 <= i < N and A[i] > A[i+1].
7+
8+ Return true if and only if the number of global inversions is equal to the number of local inversions.
9+
10+ Example 1:
11+
12+ Input: A = [1,0,2]
13+ Output: true
14+ Explanation: There is 1 global inversion, and 1 local inversion.
15+ Example 2:
16+
17+ Input: A = [1,2,0]
18+ Output: false
19+ Explanation: There are 2 global inversions, and 1 local inversion.
20+ Note:
21+
22+ A will be a permutation of [0, 1, ..., A.length - 1].
23+ A will have length in range [1, 5000].
24+ The time limit for this problem has been reduced.
25+
26+ 思路:
27+ 二分法。
28+
29+ 测试地址:
30+ https://leetcode.com/contest/weekly-contest-69/problems/global-and-local-inversions/
31+
32+ """
33+ import bisect
34+ 35+ class Solution (object ):
36+ def isIdealPermutation (self , A ):
37+ """
38+ :type A: List[int]
39+ :rtype: bool
40+ """
41+ global_inversions = []
42+ _g = 0
43+ 44+ for i in A [::- 1 ]:
45+ _g += bisect .bisect_left (global_inversions , i )
46+ bisect .insort_left (global_inversions , i )
47+ 48+ _l = 0
49+ 50+ for i in range (len (A )- 1 ):
51+ if A [i ] > A [i + 1 ]:
52+ _l += 1
53+ 54+ return _g == _l
55+
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