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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].

给定一个数组 A，我们可以将它按一个非负整数 K 进行轮调，这样可以使数组变为 A[K], A[K+1], A{K+2], ... A[A.length - 1], A[0], A[1], ..., A[K-1] 的形式。此后，任何值小于或等于其索引的项都可以记作一分。

例如，如果数组为 [2, 4, 1, 3, 0]，我们按 K = 2 进行轮调后，它将变成 [1, 3, 0, 2, 4]。这将记作 3 分，因为 1 > 0 [no points], 3 > 1 [no points], 0 <= 2 [one point], 2 <= 3 [one point], 4 <= 4 [one point]。

在所有可能的轮调中，返回我们所能得到的最高分数对应的轮调索引 K。如果有多个答案，返回满足条件的最小的索引 K。

示例 1：
输入：[2, 3, 1, 4, 0]
输出：3
解释：
下面列出了每个 K 的得分：
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
所以我们应当选择 K = 3，得分最高。 

示例 2：
输入：[1, 3, 0, 2, 4]
输出：0
解释：
A 无论怎么变化总是有 3 分。
所以我们将选择最小的 K，即 0。

提示：

• A 的长度最大为 20000。
• A[i] 的取值范围是 [0, A.length]。

 1 class Solution {
 2     func bestRotation(_ A: [Int]) -> Int {
 3         var n:Int = A.count
 4         var res:Int = 0
 5         var change:[Int] = [Int](repeating:0,count:n)
 6         for i in 0..<n
 7         {
 8             change[(i - A[i] + 1 + n) % n] -= 1
 9         }
10         for i in 1..<n
11         {
12             change[i] += change[i - 1] + 1
13             res = (change[i] > change[res]) ? i : res
14         }
15         return res
16     }
17 }