[LeetCode] Candy, Solution

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?
[Thoughts]
蛮好玩的题。感觉用dp简单点。定义Candy[i]为第i个孩子需要给的最少的糖数，
那么
Candy[i] =            Candy[i-1]+1  if ratings[i] > ratings[i-1] 递增序列，后面小孩需要的糖果是前一个小孩的糖果数+1
                           1                   if ratings[i] == ratings[i-1] 直线，按照题意，如果两个小孩rating一样多，后面的小孩可以只拿一个糖
                           Candy[i-1] –1 if ratings[i] < ratings[i-1] 递减序列。这个递推式显然是有缺陷，因为如果递减序列比较长的话，Candy[i]就可能出现负值了，负值显然是没有意义的。比如下图为例：
蓝线是rating变化曲线，数字是Candy[i]的值。基于上面递推式的解（第一行）明显是不合理的。而第二行经过调整的（红色数字），才是最优解。简单的说，就是当遇到一个波谷的时候，调整一下左边的下降序列就好了，但是要注意区分条件，上一个波峰只在有些条件下才需要更改（例一和例二的区别）。

[Code]

1:         int candy(vector<int> &ratings) {  2:            vector<int> candy(ratings.size());  3:            candy[0] = 1;  4:            int i =1;  5:            for (; i < ratings.size(); ++i)  6:            {  7:                 if (ratings[i] > ratings[i-1]) //递增  8:                 {  9:                      candy[i] = candy[i - 1] + 1;  10:                 }  11:                 if (ratings[i] == ratings[i-1]) //平行  12:                 {  13:                      candy[i] = 1;  14:                 }  15:                 if (ratings[i] < ratings[i - 1]) //递减  16:                 {  17:                      candy[i] = candy[i - 1] - 1;  18:                 }  19:                 if (i<ratings.size()-1 && ratings[i] < ratings[i-1] && ratings[i] <=ratings[i+1])  20:                      ReAdjustCandy(ratings, candy, i);  21:            }  22:            if (ratings[i-1] < ratings[i-2])  23:                 ReAdjustCandy(ratings, candy, ratings.size() - 1);  24:            int total = 0;  25:            std::for_each(candy.begin(), candy.end(), [&](int n){  26:                 total += n;  27:            });  28:            return total;  29:       }  30:       void ReAdjustCandy(vector<int>& ratings, vector<int>& candy, int startIndex)  31:       {  32:            int k = startIndex;  33:            int diff = 1 - candy[k];  34:            while (k > 0 && ratings[k - 1] > ratings[k])  35:            {  36:                 candy[k] = candy[k] + diff;  37:                 k--;  38:            }  39:            if (diff > 0) candy[k] += diff;  40:       }  

来源：https://www.cnblogs.com/codingtmd/p/5078859.html