LeetCode 347. Top K Frequent Elements

/*****************************************************************************
*
* 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:
* You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
* Your algorithm’s time complexity must be better than O(n log n), where n is the
* array’s size.
*
*****************************************************************************/
1.分析，首先此题应该用哈希表存储数字，频率
unordered_map

#include <iostream>  #include <queue>      using namespace std;    int main() {        priority_queue<int, vector<int>, less<int> >q;//使用priority_queue<int> q1;一样          for (int i = 0;i<10;i++)            q.push(i);        while (!q.empty()) {            cout << q.top() << endl;            q.pop();        }        return 0;    }// 输出9-0

因此根据上述分析可得

#include <iostream>#include<queue>#include<map>#include <unordered_map>#include <vector>#include <set>using namespace std;       vector<int> topKFrequent(vector<int>& nums, int k) {          vector<int>result;        //一，统计处频次          unordered_map<int, int> mapping;        for (auto number : nums)          //基于范围内的for循环            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;    }    //测试    int main()    {        vector<int>nums = { 1,1,1,2,2,3 };        vector<int>result;        result = topKFrequent(nums, 2);        for (auto c : result)            cout << c << endl;    }

参考博文http://blog.csdn.net/EbowTang/article/details/51317106