算法描述：

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?

Example 1:

Input: [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.

Example 2:

Input: [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
             The third child gets 1 candy because it satisfies the above two conditions.

解題思路：先從左往右掃，遇到當前小孩rating值大於前一個小孩rating值，則給該小孩分的糖置為前一個小孩糖果數加一。然後從右往左掃，遇到當前小孩rating值小於後面小孩的rating值，該小孩的糖果數為當前分配的糖果數和後面小孩分配的糖果數加一兩者之中的最大值。

    int candy(vector<int>& ratings) {
        int sum =0;
        vector<int> count(ratings.size(), 1);
        for(int i=1; i < ratings.size(); i++){
            
 
if(ratings[i] > ratings[i-1]) count[i]=count[i-1]+1;
        }
        for(int i=ratings.size()-2; i >=0; i--){
            if(ratings[i] > ratings[i+1]) count[i] = max(count[i],count[i+1]+1);
        }
        for(int i=0; i < ratings.size(); i++) sum += count[i];
        return sum;
    }

LeetCode-135-Candy