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# 347.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.

## 代码

### Approach 1: Heap

Complexity Analysis
• Time complexity : O(N_log(_k)). The complexity of `Counter method is O(N). To build a heap and output list takes O(N_log(_k)). Hence the overall complexity of the algorithm is O(N+N_log(_k)=O(N_log(_k).
• Space complexity : O(N) to store the hash map.
class Solution {
public List<Integer> topKFrequent(int[] nums, int k) {
HashMap<Integer, Integer> count = new HashMap();
for (int n : nums) {
count.put(n, count.getOrDefault(n, 0) + 1);
}
PriorityQueue<Integer> heap = new PriorityQueue<Integer>((n1, n2) -> count.get(n1) - count.get(n2));
for (int n : count.keySet()) {