[LeetCode OJ] Candy

 1 class Solution {
 2 public:
 3     int candy(vector<int> &ratings) {
 4         vector<int> candy_number(ratings.size(), 1);
 5         //不用迭代器，改用下标索引实现，这样代码看起来更加简洁
 6         for(int i=1; i<ratings.size(); ++i)
 7         {
 8            //保障当右边的rating比左边的高时，则右边分得的糖果比左边多
 9             if( ratings[i] > ratings[i-1])      //eg：输入[4 2 3 4 1],最少的分法是[2 1 2 3 1],分发9个糖果
10                 candy_number[i] = (candy_number[i] > candy_number[i-1]+1) ? candy_number[i] : candy_number[i-1]+1;
11         }
12         
13 
14         for(int i=ratings.size()-2; i>=0; --i)   
15         {
16             //保障当左边的rating比右边的高时，则左边分得的糖果比右边多
17             if( ratings[i] > ratings[i+1])
18                 candy_number[i] = (candy_number[i] > candy_number[i+1]+1) ? candy_number[i] : candy_number[i+1]+1;
19         }
20 
21         int total=0;
22         for(int i=0; i!=ratings.size(); ++i)
23             total = total + candy_number[i];
24 
25         return total;
26     }
27 };