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.

class Solution {
    public boolean isIdealPermutation(int[] A) {
        int glb = 0, loc = 0;
        for(int i = 0; i < A.length - 1; i++) {
            if(A[i] > A[i + 1]) loc++;
            for(int j = i + 1; j < A.length; j++) {
                if(A[i] > A[j]) glb++;
            }
        }
        return glb == loc;
    }
}

O(n2), TLE

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

 是local一定是global，因为global很容易，所以要返回true一定不能找到max(A[i]) > A[i + 2], 持续找，如果找不到就返回true，否则返回false

标签：775,global,int,true,Global,number,length,Inversions,local
来源： https://www.cnblogs.com/wentiliangkaihua/p/14620113.html