347. Top K Frequent Elements**

347. Top K Frequent Elements**

​​https://leetcode.com/problems/top-k-frequent-elements/​​

题目描述

Given a non-empty array of integers, return the ​​k​​ most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:

Input: nums = [1], k = 1
Output: [1]

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.

C++ 实现 1

哈希表统计个数, 最小堆取 TopK.

class Solution {
private:
    struct Comp {
        bool operator()(const pair<int, int> &p, const pair<int, int> &q) {
            return p.second > q.second;
        }
    };
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> record;
        for (auto &n : nums) record[n] ++;
        priority_queue<pair<int, int>, vector<pair<int, int>>, Comp> q;
        for (auto &p : record) {
            q.push(p);
            if (q.size() > k) q.pop();
        }
        vector<int> res;
        while (!q.empty()) {
            res.push_back(q.top().first);
            q.pop();
        }
        return res;
    }
};

C++ 实现 2

哈希表统计个数, 最大堆取 TopK.

class Solution {
private:
    struct Comp {
        bool operator() (const pair<int, int> &p, const pair<int, int> &q) {
            return p.second < q.second;
        }
    };
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> record;
        for (auto &n : nums) record[n] ++;
        priority_queue<pair<int, int>, vector<pair<int, int>>, Comp> q;
        for (auto &p : record) q.push(p);
        vector<int> res;
        while (k--) {
            auto p = q.top();
            q.pop();
            res.push_back(p.first);
        }
        return res;
    }
};