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 == 0 || len == 1)

            return len;

        int *c = new int[len];

        c[0] = 1;

        for(int i = 1; i < len; i++)

            if(ratings[i] > ratings[i-1])

                c[i] = c[i-1] + 1;

            else

                c[i] = 1;

        int minCandy = c[len-1];

        for(int i = len-2; i >= 0; i--)

        {

            if(ratings[i] > ratings[i+1])

                c[i] = max(c[i], c[i+1] + 1);

            minCandy += c[i];            

        }

        return minCandy;

        

    }

};



int main(void)

{

    int ratings[] = {5,8,2,4,9,5,4};

    int len = sizeof(ratings) / sizeof(ratings[0]);

    vector<int> ratVec(ratings, ratings+len);

    Solution solution;

    int ret = solution.candy(ratVec);

    cout<<ret<<endl;

    return 0;

}