Hard-题目34：135. Candy

题目原文：
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?
题目大意：
有N个小朋友站成一片，每个小朋友有一个评分。
现在要给这N个小朋友发糖，要求每个小朋友至少一个糖，且相邻两个小朋友中，评分高者要比评分低者得到的糖多。
请问至少需要准备多少颗糖？
题目分析：
使用贪心法来做，首先给每个小朋友一个糖，然后从左到右扫描数组，如果i号小朋友的评分高于i-1号，则给i号小朋友的糖是i-1号小朋友+1，再从右到左扫描数组，如果i号小朋友比i+1号小朋友评分高，则给i号小朋友的糖数是i+1号小朋友和i-1号小朋友的糖数最大值+1.
源码：（language：cpp）

class Solution {
public:
  int candy(vector<int> &ratings)
 {
     int size=ratings.size();
     if(size<=1)
         return size;
     vector<int> num(size,1);
     for (int i = 1; i < size; i++)
     {
         if(ratings[i]>ratings[i-1])
             num[i]=num[i-1]+1;
     }
     for (int i= size-1; i>0 ; i--)
     {
         if(ratings[i-1]>ratings[i])
             num[i-1]=max(num[i]+1,num[i-1]);
     }
     int result=0;
     for (int i = 0; i < size; i++)
     {
         result+=num[i];
        // cout<<num[i]<<" ";
     }
     return result;
 }
};

成绩：
40ms,22.09%,40ms,65.85%
Cmershen的碎碎念：
这道题好像今年alibaba的实习生机试里面考了原题，因此leetcode应重视，如果公司出题人懒，就直接从leetcode里面找题了。