347. Top K Frequent Elements

转载于 2018-04-02 08:46:52 发布 · 113 阅读

本文介绍了一种高效查找数组中出现频率最高的k个整数的方法。使用C++实现，通过哈希表统计每个元素的出现次数，并利用优先队列选出前k个高频元素。此算法的时间复杂度优于O(n log n)，适用于大数据量处理。

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.

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        
        if(nums.size()==0)
            return vector<int>();
        unordered_map<int, int> nums_count;
        for(int i=0; i<nums.size(); i++)
        {
            if(nums_count.count(nums[i]))
            {
                nums_count[nums[i]]+=1;
            }else
            {
                nums_count[nums[i]]=1;
            }
        }
        
        vector<int> res;
        priority_queue<pair<int,int>> pq; 
   
        for(auto it = nums_count.begin(); it != nums_count.end(); it++)
        {
            pq.push(make_pair(it->second, it->first));
            if(pq.size()>(int)nums_count.size()-k)
            {
                res.push_back(pq.top().second);
                pq.pop();
            }
        }
        return res;
    }
};