347. 前 K 个高频元素 - 力扣（LeetCode）

# Description

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Example 1:

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

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

Constraints:

1 <= nums.length <= 105
k is in the range [1, the number of unique elements in the array].
It is guaranteed that the answer is unique.

Follow up: Your algorithm's time complexity must be better than O(n log n), where n is the array's size.

# Code

用 min heap 的方法
time complexity O (n log n)

	class Solution {
	public int[] topKFrequent(int[] nums, int k) {

	Map<Integer, Integer> hm = new HashMap<>();
	for (int num: nums) {
	hm.put(num, hm.getOrDefault(num, 0) + 1);
	}

	Queue<int[]> queue = new PriorityQueue<int[]>(new Comparator<int[]>(){
	public int compare(int[] x, int[] y) {
	return x[1] - y[1];
	}
	});

	for (int key : hm.keySet()) {
	queue.offer(new int[]{key, hm.get(key)});
	if (queue.size() > k) {
	queue.poll();
	}
	}

	int[] res = new int[k];

	for(int i = k - 1; i >= 0; i--){
	res[i] = queue.poll()[0];
	}
	return res;
	}
	}

Reference：
O (n) 的桶排解法
347. 前 K 个高频元素 - 前 K 个高频元素 - 力扣（LeetCode）