題目：
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?
解題思路：
進行兩遍掃描，第一次從前往后掃描，若后面的child的rating比前面高，則多給一個糖果，否則，給他一個糖果。第二遍從后往前掃描，給的糖果數為candy[i] = max(candy[i], candy[i + 1] + 1);
代碼如下：
class Solution { public:int candy(vector<int> &ratings) {if (ratings.empty()) return 0;vector<int> candy(ratings.size());candy[0] = 1; //the first children, give he 1 candy//從前往后掃描for (int i = 1; i < ratings.size(); i++) {if (ratings[i] > ratings[i - 1]) {candy[i] = candy[i - 1] + 1;}else {candy[i] = 1;}}//從后向前掃描candy[ratings.size() - 1] = max(candy[ratings.size() - 1], 1);int ans = candy[ratings.size() - 1];for (int i = ratings.size() - 2; i >= 0; i--) {if (ratings[i] > ratings[i + 1]) {candy[i] = max(candy[i], candy[i + 1] + 1);}ans += candy[i];}return ans;} };
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