标签:point xpl inpu return ted att trie core etc
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]
.
s
[LeetCode] Smallest Rotation with Highest Score 得到最高分的最小旋转
标签:point xpl inpu return ted att trie core etc
原文地址:https://www.cnblogs.com/grandyang/p/9272921.html