LeetCode135：Candy

最新推荐文章于 2021-06-28 16:10:01 发布

原创最新推荐文章于 2021-06-28 16:10:01 发布 · 182 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

186 篇文章

订阅专栏

本文探讨了一个有趣的问题：如何最小化分配给孩子们的糖果数量，同时确保评分较高的孩子得到的糖果比邻居多。通过两次扫描算法，文章详细解释了实现这一目标的具体步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

There are N children standing in a line. Each child is assigned a rating value.

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.

What is the minimum candies you must give?

Example 1:

Input: [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: [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.

LeetCode：链接

初始化所有孩子糖果为1颗，接下来从左到右扫描进行“修正”。

假设我们只考虑左边孩子的情况，如果当前孩子大于左边孩子，那么糖果要在左边孩子基础上+1.

这样扫描一遍以后，我们可以保证，对于任意的 i > 0 有 ans[i]>ans[i-1] ( if ratings[i] >rating[i-1])

接下来从右向左扫描，如果当前孩子大于右边的孩子，并且糖果数少于右边的，要在右边的孩子基础上+1

总结如下：第一次扫描保证了每个比左边高分的孩子比左边的糖果多，第二次则是右边。

class Solution(object):
    def candy(self, ratings):
        """
        :type ratings: List[int]
        :rtype: int
        """
        if len(ratings) <= 1:
            return len(ratings)
        n = len(ratings)
        ans = [1] * n
        for i in range(1, n):
            if ratings[i] > ratings[i-1]:
                ans[i] = ans[i-1] + 1

        for i in range(n-2, -1, -1):
            if ratings[i] > ratings[i+1] and ans[i] <= ans[i+1]:
                ans[i] = ans[i+1] + 1

        return sum(ans)