leetcode || 135、Candy

最新推荐文章于 2025-12-06 07:43:37 发布

原创最新推荐文章于 2025-12-06 07:43:37 发布 · 713 阅读

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本文深入探讨了一个看似简单的儿童糖果分配问题，揭示了贪心算法在处理递减序列时的局限性，并提供了一种解决策略以确保最小糖果数量的分配公平性。通过案例分析和代码实现，读者将了解如何在不牺牲公平性的前提下优化资源分配。

problem：

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?

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Greedy

题意：给小盆友发糖果，属性高的糖果要更多

thinking：

（1）一道简单的贪心题目，为何通过率这么低？？题目中根本没提属性相等时怎么发糖果。又是一道不严谨的题目。提交发现，

122的小盆友发了4个糖果，说明，第三个小盆友发了1个糖果，换做你是小盆友，你会高兴吗...

（2）撇开相等的情况不说，这道题只要处理好递减的情况就行了。递增或者乱序好处理些

code：

class Solution{
    public:
int candy(vector<int> &ratings) {
    int Total = 0;    /// Total candies
    int length = 0;  /// Continuous descending length of rate
    int nPreCanCnt = 1; /// Previous child's candy count
    int beforeDenc = nPreCanCnt;
    if(ratings.begin() != ratings.end())
    {
        Total++; //Counting the first child's candy (1).
        for(vector<int>::iterator i = ratings.begin()+1; i!= ratings.end(); i++)
        {
            if(*i < *(i-1))
            {
                length++;
                if(beforeDenc <= length)
                {
                    Total++;
                }
                Total += length;
                nPreCanCnt = 1;    //This step is important, it ensures that once we leave the decending sequence, candy number start from 1
            }
            else
            {
                int curCanCnt = 0;
                if(*i > *(i-1))
                { 
                    curCanCnt = (nPreCanCnt + 1);
                }
                else
                {
                    curCanCnt = 1;
                }
                Total += curCanCnt;
                nPreCanCnt = curCanCnt;
                length = 0;    //reset length of decending sequence
                beforeDenc = curCanCnt;
            }
        }
    }
    return Total;
}
};