[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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阐述了如何在儿童排队并赋予评分的情况下，通过两次扫描分配糖果以满足每个孩子至少获得一颗糖果且评分更高的孩子得到更多糖果的需求，最终确定分配糖果的最小数量。

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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?

例子：

input ：　　　 1 1 7 6 5 9 7

第一次扫描：　　1 1 2 1 1 2 1

第二次扫描：　　1 1 3 2 1 2 1

 1 class Solution {
 2 public:
 3     int candy(vector<int> &ratings) {
 4         int res = 0;
 5         int n = ratings.size();
 6         if (n == 0) {
 7             return res;
 8         }
 9         int *t = new int[n];
10         for (int i = 0; i < n; ++i) {
11             t[i] = 1;
12         }
13         //从左向右扫描  ，保证当前比前一个多
14         for (int i = 1; i < n; ++i) {
15             if (ratings[i] > ratings[i-1]) {
16                 t[i] = t[i-1] + 1;
17             }
18         }
19         //从右向左扫描， 保证当前比后一个多     
20         for(int i = n-2;i >= 0;i--)
21         {             
22             //如果第i个的小孩的权值比i+1个小孩高,但是糖的数目却小或者相等，那么i的糖数目等于i+1的糖数目+1。             
23             if(ratings[i] > ratings[i+1] && t[i] <= t[i+1])
24             {                 
25                 t[i] = t[i+1] + 1;            
26             }      
27         }
28         for (int i = 0; i < n; ++i) {
29             res += t[i];
30         }
31         return res;
32     }
33 };