leetcode题解-347. Top K Frequent Elements

题目：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.

题目很简单，求出数组中出现频率最高的k个数即可。思路也很清晰，首先可以遍历出数组中所有数及其出现次数并将其保存在HashMap中。然后将map转化为以出现次数为索引，数为值的数组。最后再反向遍历出出现次数最大的k个数即可、代码入下，击败了70%多的用户：

    public List<Integer> topKFrequent1(int[] nums, int k) {        int n = nums.length;        HashMap<Integer, Integer> h = new HashMap();        for (int i : nums)            if (h.containsKey(i))                h.put(i, h.get(i) + 1);            else                h.put(i, 1);        List<Integer>[] fc = new ArrayList[n + 1];        for (int i : h.keySet()) {            int f = h.get(i);       //System.out.println(f + " times of " + i);            if (fc[f] == null) fc[f] = new ArrayList();            fc[f].add(i);        }        List<Integer> ans = new ArrayList();        for (int i = n, j = 0; k > 0; k--) {            for (; fc[i] == null || j == fc[i].size(); j = 0, i--);            ans.add(fc[i].get(j++));        }        return ans;    }

此外还可以使用优先级队列和树等结构来完成此题，但因为使用较为复杂的数据结构，所以效果并不好。代码如下所示：

使用队列  public List<Integer> topKFrequent2(int[] nums, int k) {        Map<Integer, Integer> freq = new HashMap<>();        for (int num : nums) {            freq.put(num, freq.getOrDefault(num, 0) + 1);        }        PriorityQueue<Map.Entry<Integer, Integer>> pq = new PriorityQueue<>((o1, o2) -> o1.getValue() - o2.getValue());        for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {            if (pq.size() < k) {                pq.offer(entry);            } else if (entry.getValue() > pq.peek().getValue()) {                pq.poll();                pq.offer(entry);            }        }        List<Integer> result = new ArrayList<>();        for (Map.Entry<Integer, Integer> entry : pq) {            result.add(entry.getKey());        }        return result;    }