135. Candy(Array; Greedy)

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?

思路：neighbor=>元素值与前、后元素有关=>

第一次前向扫描：保证high rate元素比左邻居糖多；

第二次后向扫描：保证high rate元素比右邻居糖多；

class Solution {
public:
    int candy(vector<int> &ratings) {
        vector<int> new_ratings(ratings.size(), );
        for(int i = ; i < ratings.size(); i++) //for the first time, scan from beginning
        {
            if(ratings[i] > ratings[i-])
            {
                new_ratings[i] = new_ratings[i-]+;
            }

        }
        for(int i=ratings.size()-;i>=;i--){  //for the second time, scan from the end
            if(ratings[i]>ratings[i+]&&new_ratings[i]<=new_ratings[i+]){
                new_ratings[i]=new_ratings[i+]+;
            }
        }
        int sum=;
        for(int i=;i<new_ratings.size();i++){
            sum+=new_ratings[i];
        }  

        return sum;
    }
};