Sunday, October 16, 2016

Top K Frequent Elements

Given a non-empty array of integers, return the k most frequent elements.
For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2].
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.

Use a map to track frequency of each element. Then add Map.Entry to a PriorityQueue ordered by frequency. Now output k most frequent elements.


public List topKFrequent(int[] nums, int k) {
        List<Integer> rst = new ArrayList<>(k);
        PriorityQueue<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>(new Comparator>() {
            @Override
            public int compare(Map.Entry<Integer, Integer> first, Map.Entry<Integer, Integer> second) {
                return second.getValue() - first.getValue();
            }
        });
        Map<Integer, Integer> frequency = new HashMap<>();
        for (int n : nums) {
            frequency.put(n, frequency.containsKey(n) ? frequency.get(n) + 1 : 1);
        }
        for (Map.Entry<Integer, Integer> entry : frequency.entrySet()) {
            queue.add(entry);
        }
        int size = 0;
        while (size < k) {
            rst.add(queue.poll().getKey());
            size++;
        }
        return rst;
    }


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1 comment:

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