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.

思路：

Key insights:

• every local inversion is also a global inversion
• so “local inversions == global inversions” can be interpreted as “there are only local inversions”
• if there are only local inversions, the array will be sorted after making all local inversions
• if there are inversions that are not local, the array won’t be sorted after making all local inversions

class Solution {
    public boolean isIdealPermutation(int[] A) {
        for(int i = 0;i < A.length-1;){
            if(A[i]>A[i+1]){
                int temp = A[i];
                A[i] =A[i+1];
                A[i+1] = temp;
                i += 2;
            }else{
                i++;
            }
        }
        for(int i = 0;i < A.length -1; i++){
            if(A[i]>A[i+1]){
                return false;
            }
        }
        return true;
    }
}

public class Main {
    public static void main(String[] args){
        Solution solution = new Solution();
        int[] A = {1,2,0};
        solution.isIdealPermutation(A);
    }
}

原文地址：https://www.cnblogs.com/lxk2010012997/p/8419459.html