Candy

最新推荐文章于 2024-09-18 07:45:00 发布

cgdong2014

最新推荐文章于 2024-09-18 07:45:00 发布

阅读量259

点赞数

CC 4.0 BY-SA版权

分类专栏： Remember LeetCode

本文链接：https://blog.youkuaiyun.com/cgdong2014/article/details/39470427

LeetCode 同时被 2 个专栏收录

139 篇文章

订阅专栏

Remember

35 篇文章

订阅专栏

本文讨论了如何在给定儿童评级的基础上，通过从前到后和从后到前的遍历策略，确保每个儿童至少获得一颗糖果，并且根据评级给予更多糖果，从而计算出分配糖果所需的最小数量。

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?

思路：先从前往后，如果rating比前面的那个大，就在前面那个基础上加1，否则为1.

再从后往前，如果rating 比后面那个大，但是candy 没有比后面那个大，就要在后面那个基础上加1.

易错点：记住后面往前时候， if(ratings[i] > ratings[i + 1] && candys[i] <= candys[i + 1])才能 candys[i] = candys[i + 1] + 1;

public class Solution {
    public int candy(int[] ratings) {
        int n = ratings.length;
        int[] candys = new int[n];
        int ret = 0;
        candys[0] = 1;
        for(int i = 1; i < n; i++){
            if(ratings[i] > ratings[i - 1]){
                candys[i] = candys[i - 1] + 1;
            }else{
                candys[i] = 1;
            }
        }
        
        ret = candys[n - 1];
        for(int i = n - 2; i >= 0; i--){
            if(ratings[i] > ratings[i + 1] && candys[i] <= candys[i + 1]){
                candys[i] = candys[i + 1] + 1;
            }
            ret += candys[i];
        }
        return ret;
    }
}