347. Top K Frequent Elements

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题目

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

示例1:

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

示例2：

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

提示:

  1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
  2. Your algorithm’s time complexity must be better than O(n log n), where n is the array’s size.

解法

解法一：

构造一个大顶堆，取K个即可。

JAVA

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class Node {
		private int value;
		
		private int count;
		
		public Node(int value, int count) {
			this.value = value;
			this.count = count;
		}

		public int getValue() {
			return value;
		}

		public void setValue(int value) {
			this.value = value;
		}

		public int getCount() {
			return count;
		}

		public void setCount(int count) {
			this.count = count;
		}
	}
	
	public List<Integer> topKFrequent(int[] nums, int k) {
		Map<Integer, Integer> values = new HashMap<Integer, Integer>();
		
		for (int num : nums) {
			if (values.containsKey(num)) {
				values.put(num, values.get(num) + 1);
			} else {
				values.put(num, 1);
			}
		}
		
		PriorityQueue<Node> nodes = new PriorityQueue<>(values.size(), (a, b) -> b.getCount() - a.getCount());
		for (Map.Entry<Integer, Integer> entry : values.entrySet()) {
			Node node = new Node(entry.getKey(), entry.getValue());
			nodes.add(node);
		}
		
		List<Integer> result = new ArrayList<Integer>(k);
		while (k > 0 && !nodes.isEmpty()) {
			result.add(nodes.poll().getValue());
			k--;
		}
		
        return result;
    }
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