[Swift]LeetCode798. 得分最高的最小轮调 | Smallest Rotation with Highest Score

概述 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

 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 [1,2,4].  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: [2,0]Output: 3Explanation:  scores for each K are Listed below: K = 0,A = [2,0],score 2K = 1,A = [3,2],score 3K = 2,A = [1,3],score 3K = 3,A = [4,1],score 4K = 4,A = [0,4],score 3

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

Example 2:input: [1,4]Output: 0Explanation:  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] 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 [2,0]，我们按 K = 2 进行轮调后，它将变成 [1,4]。这将记作 3 分，因为 1 > 0 [no points],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：[2,0]输出：3解释：下面列出了每个 K 的得分：K = 0,score 3所以我们应当选择 K = 3，得分最高。 

示例 2：输入：[1,4]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]。 Runtime: 196 ms Memory Usage: 18.9 MB

 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..<n11         {12             change[i] += change[i - 1] + 113             res = (change[i] > change[res]) ? i : res14         }15         return res16     }17 }

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