# LeetCode 775. Global and Local Inversions

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.

A[i]必定只能在它本应在的位置或这个位置的左边一个或者右边一个，不然必定会把比它大的两个数字挤到前面或者把比它小的两个数字挤到后面，因为题目中已经说明它所有的数字是0～n-1的一个排列，那么A[i]-i的绝对值必定<=1，否则就不满足条件～

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