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?

思路：分糖果问题，用贪心算法。我们用num[i]表示第i个孩子的糖果数目，如果i+1孩子比i孩子的ratings大，则i+1的孩子糖果数目为i孩子糖果数目+1.如果i孩子比i+1孩子ratings大，怎么办，等从左往右开始遍历结束，我们从右往左开始，把之前忽略的情况给补上，就可以了。时间复杂度为O(n),空间复杂度为O(n)。

class Solution {

public:

    int candy(vector<int> &ratings) {

        int n=ratings.size();

        if(n<=0)

            return 0;

        vector<int> num(n,1);

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

        {

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

            {

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

            }

        }

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

        {

            if(ratings[i]>ratings[i+1] && num[i]<=num[i+1])

            {

                num[i]=num[i+1]+1;

            }

        }

        int sum=0;

        for(int i=0;i<n;i++)

        {

            sum+=num[i];

        }

        return sum;

    }

};