Leetcode 347. Top K Frequent Elements--python1行解法，Java 11ms解法

题目地址：

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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.

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python一行解法：

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        return [key for key, value in collections.Counter(nums).most_common(k)]

java官方解法：

class Solution {
  public List<Integer> topKFrequent(int[] nums, int k) {
    // build hash map : character and how often it appears
    HashMap<Integer, Integer> count = new HashMap();
    for (int n: nums) {
      count.put(n, count.getOrDefault(n, 0) + 1);
    }

    // init heap 'the less frequent element first'
    PriorityQueue<Integer> heap =
            new PriorityQueue<Integer>((n1, n2) -> count.get(n1) - count.get(n2));

    // keep k top frequent elements in the heap
    for (int n: count.keySet()) {
      heap.add(n);
      if (heap.size() > k)
        heap.poll();
    }

    // build output list
    List<Integer> top_k = new LinkedList();
    while (!heap.isEmpty())
      top_k.add(heap.poll());
    Collections.reverse(top_k);
    return top_k;
  }
}

有一个比官方解答快3倍的Java解法：

class Solution {
public List<Integer> topKFrequent(int[] nums, int k) {

	List<Integer>[] bucket = new List[nums.length + 1];
	Map<Integer, Integer> frequencyMap = new HashMap<Integer, Integer>();

	for (int n : nums) {
		frequencyMap.put(n, frequencyMap.getOrDefault(n, 0) + 1);
	}

	for (int key : frequencyMap.keySet()) {
		int frequency = frequencyMap.get(key);
		if (bucket[frequency] == null) {
			bucket[frequency] = new ArrayList<>();
		}
		bucket[frequency].add(key);
	}

	List<Integer> res = new ArrayList<>();

	for (int pos = bucket.length - 1; pos >= 0 && res.size() < k; pos--) {
		if (bucket[pos] != null) {
			res.addAll(bucket[pos]);
		}
	}
	return res;
}
}