題目
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;
}
}
