Leetcode 135. 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?

思路：

这题本身可以用贪心法来做，我们用candy[n]表示每个孩子的糖果数，遍历过程中，如果孩子i+1的rate大于孩子i 的rate，那么当前最好的选择自然是：给孩子i+1的糖果数=给孩子i的糖果数+1

如果孩子i+1的rate小于等于孩子i 的rate咋整？这个时候就不大好办了，因为我们不知道当前最好的选择是给孩子i+1多少糖果。

解决方法是：暂时不处理这种情况。等数组遍历完了，我们再一次从尾到头遍历数组，这回逆过来贪心，就可以处理之前略过的孩子。

最后累加candy[n]即得到最小糖果数。

代码：

class Solution {public:    int candy(vector<int>& ratings) {        int num = ratings.size();        vector<int> res(num, 1);        for(int i = 1; i < num; ++ i){            if(ratings[i] > ratings[i-1])                res[i] = res[i-1] + 1;        }        for(int i = num-1; i >0; --i){            if(ratings[i-1] > ratings[i])                res[i-1] = max(res[i]+1, res[i-1]);        }        int sum = 0;        for(int i = 0; i < num; ++ i){            sum += res[i];        }        return sum;    }};