There are n children standing in a line. Each child is assigned a rating value given in the integer array ratings.

You are giving candies to these children subjected to the following requirements:

• Each child must have at least one candy.
• Children with a higher rating get more candies than their neighbors.

Return the minimum number of candies you need to have to distribute the candies to the children.

Example 1:

Input: ratings = [1,0,2] Output: 5 Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively. 

Example 2:

Input: ratings = [1,2,2] Output: 4 Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively. The third child gets 1 candy because it satisfies the above two conditions. 

Constraints:

• n == ratings.length
• 1 <= n <= 2 * 10^4
• 0 <= ratings[i] <= 2 * 10^4

老师想给孩子们分发糖果，有 N 个孩子站成了一条直线，老师会根据每个孩子的表现，预先给他们评分。你需要按照以下要求，帮助老师给这些孩子分发糖果：

• 每个孩子至少分配到 1 个糖果。
• 评分更高的孩子必须比他两侧的邻位孩子获得更多的糖果。

那么这样下来，老师至少需要准备多少颗糖果呢？

• 本题的突破口在于，评分更高的孩子必须比他两侧的邻位孩子获得更多的糖果，这句话。这个规则可以理解为 2 条规则，想象成按身高排队，站在下标为 0 的地方往后“看”，评分高即为个子高的，应该比前面个子矮(评分低)的分到糖果多；站在下标为 n - 1 的地方往后“看”，评分高即为个子高的，同样应该比前面个子矮(评分低)的分到糖果多。你可能会有疑问，规则都是一样的，为什么会出现至少需要多少糖果呢？因为可能出现评分一样高的同学。扫描数组两次，处理出每一个学生分别满足左规则或右规则时，最少需要被分得的糖果数量。每个人最终分得的糖果数量即为这两个数量的最大值。两次遍历结束，将所有糖果累加起来即为至少需要准备的糖果数。由于每个人至少分配到 1 个糖果，所以每个人糖果数再加一。

package leetcode funccandy(ratings []int) int { candies:=make([]int, len(ratings)) fori:=1; i<len(ratings); i++ { ifratings[i] >ratings[i-1] { candies[i] +=candies[i-1] +1 } } fori:=len(ratings) -2; i>=0; i-- { ifratings[i] >ratings[i+1] &&candies[i] <=candies[i+1] { candies[i] =candies[i+1] +1 } } total:=0for_, candy:=rangecandies { total+=candy+1 } returntotal }