Given an array A, we may rotate it by a non-negative integer K so that the array becomes A[K], A[K+1], A{K+2], ... A[A.length - 1], A[0], A[1], ..., A[K-1].  Afterward, any entries that are less than or equal to their index are worth 1 point. 

For example, if we have [2, 4, 1, 3, 0], and we rotate by K = 2, it becomes [1, 3, 0, 2, 4].  This is worth 3 points because 1 > 0 [no points], 3 > 1 [no points], 0 <= 2 [one point], 2 <= 3 [one point], 4 <= 4 [one point].

Over all possible rotations, return the rotation index K that corresponds to the highest score we could receive.  If there are multiple answers, return the smallest such index K.

Example 1:
Input: [2, 3, 1, 4, 0]
Output: 3
Explanation:  
Scores for each K are listed below: 
K = 0,  A = [2,3,1,4,0],    score 2
K = 1,  A = [3,1,4,0,2],    score 3
K = 2,  A = [1,4,0,2,3],    score 3
K = 3,  A = [4,0,2,3,1],    score 4
K = 4,  A = [0,2,3,1,4],    score 3

So we should choose K = 3, which has the highest score.

Example 2:
Input: [1, 3, 0, 2, 4]
Output: 0
Explanation:  A will always have 3 points no matter how it shifts.
So we will choose the smallest K, which is 0.

Note:

• A will have length at most 20000.
• A[i] will be in the range [0, A.length].

数组旋转后，找出对应得分最高的旋转数。

方法一：直接求解，会TLE，程序有下所示：

class Solution {
    public int bestRotation(int[] A) {
        int[] B = new int[2*A.length];
        for (int i = 0; i < A.length; ++ i){
            B[i] = A[i];
        }
        int index = 0;
        int res = 0, max = Integer.MIN_VALUE;
        for (int i = 0; i < A.length; ++ i){
            int cnt = 0;
            for (int j = index; j < A.length + index; ++ j){
                if (B[j] <= j - index){
                    cnt ++;
                }
            }
            if (max < cnt){
                max = cnt;
                res = i;
            }
            B[A.length + index] = A[i];
            index ++;
        }
        return res;
    }
}

方法二：时间复杂度O（n），待补充...