candy——动态规划

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?

 1 int candy(vector<int> &ratings) {
 2         //题意:N个孩子站成一排,每个孩子分配一个分值。给这些孩子派发糖果,满足如下要求:
 3         //每个孩子至少一个
 4         //分值更高的孩子比他的相邻位的孩子获得更多的糖果
 5         //求至少分发多少糖果?
 6         int len=ratings.size();
 7         if(len==1) return 1;
 8           
 9         int sum=0;
10         vector<int> v(len,1);//初始将每个孩子的糖果数都设为1
11           
12         //从左向右扫描,保证一个方向上分数更大的糖果更多
13         for(int i=1;i<len;i++){
14             if(ratings[i] > ratings[i-1])
15                 v[i]=v[i-1]+1;
16         }
17         //从右向左扫描,保证另一个方向上分数更大的糖果更多
18         for(int i=len-2;i>=0;i--){
19             if(ratings[i] > ratings[i+1] && v[i] <= v[i+1])
20                 v[i]=v[i+1]+1;
21         }
22           
23         for(int i=0;i<len;i++){
24             sum+=v[i];
25         }
26         return sum;
27     }
28 };
原文地址:https://www.cnblogs.com/zl1991/p/6964113.html