算法练习13：leetcode习题135. Candy

最新推荐文章于 2021-08-22 22:16:12 发布

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最新推荐文章于 2021-08-22 22:16:12 发布

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本文链接：https://blog.youkuaiyun.com/m0_37779608/article/details/84572697

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本文探讨了一个经典的贪心算法问题：糖果分配。在确保每个孩子至少获得一颗糖果的前提下，高评级的孩子必须比邻居获得更多的糖果。文章通过具体实例解释了如何在满足条件的情况下，以最小数量的糖果完成分配，同时提供了详细的C++实现代码。

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题目

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.

算法思路

本周复习贪心算法的思想，本题也是比较典型地用到贪心算法的题目。

主要要求是相邻的两个小孩如果等级不等，那么高等级的孩子分到的糖果要比低等级的孩子多，而当等级相等时则只需保证至少有一颗就行。根据相邻这个特点，我们可以分别向左和向右扫一遍，这样就足够检查相邻孩子的等级关系。当遇到逐渐递增的等级时，分到的糖果也一颗一颗递增，直到到达局部的峰顶。

C++代码

class Solution {
public:
    int candy(vector<int>& ratings) {
        if (ratings.size() == 0)
        {
            return 0;
        }

        int count[ratings.size()];
        count[0] = 1;

        for (int i = 1; i < ratings.size(); ++i)
        {
            if (ratings[i] > ratings[i-1])
            {
                count[i] = count[i-1] + 1;
            }
            else{
                count[i] = 1;
            }
        }

        int result = count[ratings.size()-1];
        for (int i = ratings.size()-2; i >= 0; --i)
        {
            if (ratings[i] > ratings[i+1] && count[i] < count[i+1]+1)
            {
                count[i] = count[i+1]+1;
            }
            result += count[i];
        }

        return result;
    }
};