347. Top K Frequent Elements

Total Accepted: 2605  Total Submissions: 5933  Difficulty: Medium

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.

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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