0135. Candy

135. Candy #

题目 #

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

题目大意 #

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?

解题思路 #

  • 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。

代码 #

package leetcode

func candy(ratings []int) int {
	candies := make([]int, len(ratings))
	for i := 1; i < len(ratings); i++ {
		if ratings[i] > ratings[i-1] {
			candies[i] += candies[i-1] + 1
		}
	}
	for i := len(ratings) - 2; i >= 0; i-- {
		if ratings[i] > ratings[i+1] && candies[i] <= candies[i+1] {
			candies[i] = candies[i+1] + 1
		}
	}
	total := 0
	for _, candy := range candies {
		total += candy + 1
	}
	return total
}

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