Top K Frequent Elements

题目

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.

思路

先统计每个元素出现次数，然后取前k个出现最多元素
一般用hash表统计，用最大堆取前k个

cpp

class Solution {public:    vector<int> topKFrequent(vector<int>& nums, int k) {        vector<pair<int, int>> numCount;        sort(nums.begin(),nums.end());        for (auto i : nums)        {            if (numCount.empty() || numCount.back().first != i)                numCount.push_back({ i, 1 });            else//numCount.back.first==i                numCount.back().second++;        }        sort(numCount.begin(), numCount.end(), [](pair<int,int> &left,pair<int,int> &right){            return left.second > right.second;        });        vector<int> ret;        for (size_t i = 0; i < k; ++i)            ret.push_back(numCount[i].first);        return ret;    }};//hash表加最大堆class Solution {public:    vector<int> topKFrequent(vector<int>& nums, int k) {        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;        unordered_map<int, int> cnt;        for (auto num : nums) cnt[num]++;        for (auto kv : cnt) {            pq.push({kv.second, kv.first});            while (pq.size() > k) pq.pop();        }        vector<int> res;        while (!pq.empty()) {            res.push_back(pq.top().second);            pq.pop();        }        return res;    }};//hash加bucketclass Solution {public:    vector<int> topKFrequent(vector<int>& nums, int k) {        vector<int> res;        if (!nums.size()) return res;        unordered_map<int, int> cnt;        for (auto num : nums) cnt[num]++;        vector<vector<int>> bucket(nums.size() + 1);        for (auto kv : cnt) {            bucket[kv.second].push_back(kv.first);        }        for (int i = bucket.size() - 1; i >= 0; --i) {            for (int j = 0; j < bucket[i].size(); ++j){                res.push_back(bucket[i][j]);                if (res.size() == k) return res;            }        }        return res;    }};

python

from collections import Counterclass Solution(object):    def topKFrequent(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: List[int]        """        counter=Counter(nums).most_common(k)        return [key for key,count in counter]