Candy

问题来源

问题描述

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?

问题分析

如果将rating值连成一条曲线，这道问题就十分清晰了。我们的任务便是找出rating上升与rating下降的部分，换而言之，就是要找出rating函数中的极大值点。我们现在来思考上升和下降的问题，实际上上升与下降是可以相互转化的。我们可以从左至右来查找上升的部分，然后再从右至左查找上升的部分（即从左到右的下降部分）。针对此题，实际上，在函数值平坦的部分我们根据贪婪原则当然是都只给1个糖果，只在上升阶段逐个递增。比较从左至右与从右至左得到的糖果结果，取其大者即可，意即选取上升边较长的一边作为最终结果。

解决代码

class Solution {public:    int candy(vector<int>& ratings) {        vector<int> l2r(ratings.size(), 1);        vector<int> r2l(ratings.size(), 1);        l2r.push_back(1);        for (auto i = 1; i < ratings.size(); i++) {            if (ratings[i] > ratings[i-1]) {                l2r[i] = l2r[i-1] + 1;            }        }        r2l.push_back(1);        for (auto i = ratings.size()-1; i >= 1; i--) {            if (ratings[i] < ratings[i-1]) {                r2l[i-1] = r2l[i] + 1;            }        }        int count = 0;        for (auto i = 0; i < ratings.size(); i++) {            count += max(l2r[i], r2l[i]);        }        return count;    }};