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：

　　进行两遍扫描，第一次从前往后扫描，若后面的child的rating比前面高，则多给一个糖果，否则，给他一个糖果。第二遍从后往前扫描，给的糖果数为candy[i] = max(candy[i], candy[i + 1] + 1);

代码如下：

　　

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//从前往后扫描for (int i = 1; i < ratings.size(); i++) {if (ratings[i] > ratings[i - 1]) {candy[i] = candy[i - 1] + 1;}else {candy[i] = 1;}}//从后向前扫描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;} };

 

 

转载于:https://www.cnblogs.com/dongguangqing/p/3727164.html

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