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=> 需要前向、后向两次扫描

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(), 1);
        for(int i = 1; i < ratings.size(); i++) //for the first time, scan from beginning
        {
            if(ratings[i] > ratings[i-1]) 
            {
                new_ratings[i] = new_ratings[i-1]+1;
            }
          
        }
        for(int i=ratings.size()-2;i>=0;i--){  //for the second time, scan from the end
            if(ratings[i]>ratings[i+1]&&new_ratings[i]<=new_ratings[i+1]){  
                new_ratings[i]=new_ratings[i+1]+1;  
            }  
        }  
        int sum=0;  
        for(int i=0;i<new_ratings.size();i++){  
            sum+=new_ratings[i];  
        }  
          
        return sum;  
    }
};