Global and Local Inversions

题目
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.

答案

class Solution {
    // there is no global inversion with j > i + 1
    // basically, we want to find if there is a case where
    // A[i] > A[j], and j >= i + 2
    public boolean isIdealPermutation(int[] A) {
        if(A.length == 0) return true;
        int max = Integer.MIN_VALUE;
        for(int i = 0; i < A.length - 2; i++) {
            max = Math.max(max, A[i]);
            if(max > A[i + 2]) return false;
        }
        return true;
    }
}