# Candy

> 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?

好了,终于到了小盆友,排队领糖果的时候了,我们可是坏叔叔。

这题要求每个小孩至少要领到一颗糖果,但是高级别的小孩要比它旁边的孩子得到的糖果多(小孩的世界也有不平等了),问最少需要发多少糖果?

首先我们会给每个小朋友一颗糖果,然后从左到右,假设第i个小孩的等级比第i - 1个小孩高,那么第i的小孩的糖果数量就是第i - 1个小孩糖果数量在加一。再我们从右到左,如果第i个小孩的等级大于第i + 1个小孩的,同时第i个小孩此时的糖果数量小于第i + 1的小孩,那么第i个小孩的糖果数量就是第i + 1个小孩的糖果数量加一。

代码如下:

```c++
class Solution {
public:
int candy(vector &ratings) {
vector candys;
//首先每人发一颗糖
candys.resize(ratings.size(), 1);
//这个循环保证了右边高级别的孩子一定比左边的孩子糖果数量多
for(int i = 1; i < (int)ratings.size(); i++) {
if(ratings[i] > ratings[i - 1]) {
candys[i] = candys[i - 1] + 1;
}
}

//这个循环保证了左边高级别的孩子一定比右边的孩子糖果数量多
for(int i = (int)ratings.size() - 2; i >= 0; i--) {
if(ratings[i] > ratings[i + 1] && candys[i] <= candys[i + 1]) {
candys[i] = candys[i + 1] + 1;
}
}

int n = 0;
for(int i = 0; i < (int)candys.size(); i++) {
n += candys[i];
}
return n;
}
};
```