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?

 int candy(vector<int> &ratings) {
         //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：
         //每个孩子至少一个
         //分值更高的孩子比他的相邻位的孩子获得更多的糖果
         //求至少分发多少糖果？
         int len=ratings.size();
         if(len==) return ;

         int sum=;
         vector<int> v(len,);//初始将每个孩子的糖果数都设为1

         //从左向右扫描，保证一个方向上分数更大的糖果更多
         for(int i=;i<len;i++){
             if(ratings[i] > ratings[i-])
                 v[i]=v[i-]+;
         }
         //从右向左扫描，保证另一个方向上分数更大的糖果更多
         for(int i=len-;i>=;i--){
             if(ratings[i] > ratings[i+] && v[i] <= v[i+])
                 v[i]=v[i+]+;
         }

         for(int i=;i<len;i++){
             sum+=v[i];
         }
         return sum;
     }
 };