[LeetCode] Candy

最新推荐文章于 2023-02-28 14:31:32 发布

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最新推荐文章于 2023-02-28 14:31:32 发布

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本文探讨了如何在儿童排队时，确保每个孩子至少获得一个糖果，并且糖果数量随评级递增分配的策略。通过双扫描算法，实现从左到右和从右到左的分配过程，确保糖果分配的公平性和合理性。

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?

有一排小孩，给他们分糖果，每人至少分一个，高个比相邻的低个分的糖果多。

两次扫描，先从左往右，从k=1开始分配，遇到高个k++,遇到矮个子 k = 1;

然后从右往左扫描，k = max(1,pre) pre是之前扫描是的糖果，遇到搞个k++，遇到矮个子 k = max(1,pre);

class Solution {
public:
    int candy(vector<int> &ratings) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(ratings.size() <= 1) return ratings.size();
        
        int *num = new int[ratings.size()];
        for(int i = 0 ; i < ratings.size(); i++) num[i] = 0;
        int sum = ratings.size();
        
        for(int i = 0, k = 1; i < ratings.size(); i++) {
            if(i-1 >= 0 && ratings[i] > ratings[i-1]) {
                num[i] = (k > num[i] ? k : num[i]);
                k++;
            }else {
                k = 1;
            }
        }
        
        for(int i = ratings.size()-1, k = 1; i >= 0; i--) {
            if(i+1 < ratings.size() && ratings[i] > ratings[i+1]) {
                num[i] = (k > num[i] ? k : num[i]);
                k++;
            }else {
                k = 1;
            }
        }
        int total = ratings.size();
        for(int i = 0 ; i < ratings.size(); i++) total += num[i];
        delete []num;
        
        return total;
    }
};