LeetCode135：Candy

题目：

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?

解题思路：

遍历两遍数组即可

第一遍：如果当前元素比前一个大，则当前小孩获得的糖果数为前一个小孩糖果数+1，否则糖果数为1

第二遍：从后往前扫描，如果当前元素i的值大于i+1位置的值，则比较两者目前的糖果数，如果i小孩获得的糖果数大于第i+1个小孩获得糖果数+1，则不变，否则，将i小孩糖果数置为第i+1个小孩糖果数+1.

实现代码：

#include <iostream>
#include <vector>
using namespace std;
 
class Solution {
public:
    int candy(vector<int> &ratings) {
        int len = ratings.size();
        if(len ==  || len == )
            return len;
        int *c = new int[len];
        c[] = ;
        for(int i = ; i < len; i++)
            if(ratings[i] > ratings[i-])
                c[i] = c[i-] + ;
            else
                c[i] = ;
        int minCandy = c[len-];
        for(int i = len-; i >= ; i--)
        {
            if(ratings[i] > ratings[i+])
                c[i] = max(c[i], c[i+] + );
            minCandy += c[i];
        }
        return minCandy;
 
    }
};
 
int main(void)
{
    int ratings[] = {,,,,,,};
    int len = sizeof(ratings) / sizeof(ratings[]);
    vector<int> ratVec(ratings, ratings+len);
    Solution solution;
    int ret = solution.candy(ratVec);
    cout<<ret<<endl;
    return ;
}