【leetcode】c++求数组中出现频率最多的 k 个数

来源： https://leetcode.com/problems/top-k-frequent-elements/discuss/81624/C%2B%2B-O(n-log(n-k))-unordered_map-and-priority_queue(maxheap)-solution

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.

解题思路

利用unordered_map和priority_queue实现

#include 
#include 
vector<int> topKFrequent(vector<int>& nums, int k)
{
	//
	//key-value == 数组值-频率
	unordered_map<int, int> map;
	for (int num : nums)//简单的循环方法
	{
		map[num]++;//求出每一个数的出现的频率value，类比灰度分布图
	}

	vector<int> res;
	//
	// pair: first is frequency,  second is number
	priority_queue<pair<int, int>> pq;
	for (auto it = map.begin(); it != map.end(); it++)
	{
		pq.push(make_pair(it->second, it->first));//按频率进行排序即it->second
		if (pq.size() > (int)map.size() - k)//找到前k个频率最大的数
		{
			res.push_back(pq.top().second);
			pq.pop();
		}
	}
	return res;
}

unordered_map解释

https://www.cnblogs.com/tp-16b/p/9156810.html

priority_queue解释

https://blog.csdn.net/weixin_36888577/article/details/79937886