【leetcode】Global and Local Inversions
题目如下:
We have some permutation A of [0, 1, ..., N - 1], where N is the length of A. The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j]. The number of local inversions is the number of i with 0 <= i < N and A[i] > A[i+1]. Return true if and only if the number of global inversions is equal to the number of local inversions.
Example 1: Input: A = [1,0,2] Output: true Explanation: There is 1 global inversion, and 1 local inversion.
Example 2: Input: A = [1,2,0] Output: false Explanation: There are 2 global inversions, and 1 local inversion.
Note: A will be a permutation of [0, 1, ..., A.length - 1]. A will have length in range [1, 5000]. The time limit for this problem has been reduced.
解题思路:这是一个非常有趣的题目。我最初的想法是用一个二层的循环计算出Global的值,然后好Local比较,但是O(n^2)的复杂度显然是不行的,然后我各种优化但是得到的依旧是冷冰冰的Time Exceed Limit.考虑到题目只是要求判断Global是否会等于Local,机智的我果断放弃了计算出Global的值的尝试,开始寻找更巧妙的解法。皇天不负有心人,果然被我找到了。对应任意一个数字A[i]来说,它的Global一定是大于或者等于Local的,也就是说,在整个A中,只要找到任意一个A[i]满足在A[i+2]~A[N-1]里面有一个小于A[i]的数,最终整个数组的Global就一定会大于Local了。为什么不用考虑A[i+1]?无论A[i]和A[i+1]的大小关系如何,Global和Local的值要不分别加1,要么不变,所以不管。那么问题来了,怎么找到任意一个A[i]满足在A[i+2]~A[N-1]里面有一个小于A[i]的数呢?只需要依次遍历数组,找出以遍历元素中的最大值(记为max)和当前遍历到的元素的后面第二个元素(记为A[i+2])比较就行了,如果max > A[i+2] ,那么Global就一定会大于Local了。
代码如下:
class Solution(object): def isIdealPermutation(self, A): """ :type A: List[int] :rtype: bool """ maxV = -1 for i in range(len(A)-2): if maxV == -1 or maxV < A[i]: maxV = A[i] if maxV > A[i+2]: return False return True