Description:

There are two properties in the node student id and scores, to ensure that each student will have at least 5 points, find the average of 5 highest scores for each person.
给出一系列学生分数，求出每个学生最高的5门分数的平均分。

Example:

Given results = [[1,91],[1,92],[2,93],[2,99],[2,98],[2,97],[1,60],[1,58],[2,100],[1,61]]

Return

Link:

http://www.lintcode.com/en/problem/high-five/

解题思路：

本题重点是如果存储每个学生最高的5门的成绩，这里针对每个学生可以使用一个vector，这个vector容积为5，从第0到第4位以递减的方式储存5门成绩；当输入一个成绩时，从第0位开始比较，如果输入的成绩大于这一位置的原有成绩时，数组由这一位置整体向后挪一位，并且更新这一位置的成绩。

Time Complexity：

O(N)

完整代码：

map<int, double> highFive(vector<Record>& results) { map<int, vector<double>> record; map<int, double> output; for(Record r : results) { if(record[r.id].size() == 0) { vector<double> temp(5, 0); helper(temp, r.score); record[r.id] = temp; } else { helper(record[r.id], r.score); } } for(auto a : record) { double sum = 0; for(double d : a.second) sum += d; output[a.first] = sum/5; } return output; } void helper(vector<double>& v, int score) { for(int i = 0; i < 5; i++) { if(score <= v[i]) continue; else { for(int j = 4; j > i; j--) v[j] = v[j-1]; v[i] = score; return; } } }