Given a non-empty array of integers, return the k most frequent elements.
For example,
Given [1,1,1,2,2,3]
and k = 2, return [1,2]
.
Note:
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分析:
两个核心操作,
1,利用哈希map统计频次及其对应数字
2,根据频次(及其对应数字)建立最大堆,然后我们总是弹出堆顶就能获取当前最大频次
class Solution { public: vector<int> topKFrequent(vector<int>& nums, int k) { //一,统计处频次 unordered_map<int,int> mapping; for(int number : nums) mapping[number]++; //二,根据频次压入最大堆中 // pair<first, second>: first is frequency, second is number priority_queue<pair<int,int>> pri_que; //最大堆 for(auto it = mapping.begin(); it != mapping.end(); it++) pri_que.push(make_pair(it->second, it->first)); //三,获取结果 while(result.size() < k){ result.push_back(pri_que.top().second); pri_que.pop(); } return result; } private: vector<int> result; };
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原文地址:http://blog.csdn.net/ebowtang/article/details/51317106
原作者博客:http://blog.csdn.net/ebowtang
本博客LeetCode题解索引:http://blog.csdn.net/ebowtang/article/details/50668895