347. 找出出现频次最高的 k 个数

347. 找出出现频次最高的 k 个数
347. Top K Frequent Elements(medium)

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

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

解法1

堆处理词频顺序
1.初始化词频
2.通过 Map.Entry 进行排序
3. 弹出 k 个元素即为词频最高的 k 个数

public List<Integer> topKFrequent(int[] nums, int k) {
    Map<Integer, Integer> freqs = new HashMap<>();
    for (int num : nums)
        freqs.put(num, freqs.getOrDefault(num, 0) + 1);

    PriorityQueue<Map.Entry<Integer, Integer>> pq = new PriorityQueue<>((o1,o2) -> !o1.getValue().equals(o2.getValue()) ? Integer.compare(o1.getValue(), o2.getValue()) : Integer.compare(o1.getKey(), o2.getKey()));
    for (Map.Entry<Integer, Integer> entry : freqs.entrySet()) {
        pq.offer(entry);
        if (pq.size() == k + 1)
            pq.poll();
    }
    List<Integer> res = new ArrayList<>();
    while (!pq.isEmpty())
        res.add(pq.poll().getKey());
    return res;
}

解法2

桶排序实现, 统计词频, 之后按照词频作为索引, 存放对应词频的数

• index : frequency
• value : 4 <=> 89 <=> 32 <=>

public List<Integer> topKFrequent(int[] nums, int k) {
    Map<Integer, Integer> freqs = new HashMap<>();
    for (int val : nums)
        freqs.put(val, freqs.getOrDefault(val, 0) + 1);

    List<Integer>[] buckets = new ArrayList[nums.length + 1];
    for (Map.Entry<Integer, Integer> entry : freqs.entrySet()) {
        int freq = entry.getValue();
        if (buckets[freq] == null)
            buckets[freq] = new ArrayList<>();
        buckets[freq].add(entry.getKey());
    }

    List<Integer> topK = new ArrayList<>();
    for (int i = buckets.length - 1; i >= 0; i --) {
        if (buckets[i] == null) continue;
        if (topK.size() >= k) break;
        if (buckets[i].size() <= (k - topK.size()))
            topK.addAll(buckets[i]);
        else
            topK.addAll(buckets[i].subList(0, k - topK.size()));
    }
    return topK;
}

解法3

LinkedHashMap 处理词频顺序
按照 Entry 的 value 进行排序, 之后重新插入到 LinkedHashMap 中, 保证迭代时的有序性

public List<Integer> topKFrequent(int[] nums, int k) {
    Map<Integer, Integer> freqs = new HashMap<>();
    for (int num : nums)
        freqs.put(num, freqs.getOrDefault(num, 0) + 1);

    freqs = sortByValue(freqs);
    List<Integer> res = new ArrayList<>();
    for (Map.Entry<Integer, Integer> entry : freqs.entrySet()) {
        res.add(entry.getKey());
        if (res.size() == k)
            return res;
    }
    throw new IllegalArgumentException();
}

private Map<Integer, Integer> sortByValue(Map<Integer, Integer> map) {
    List<Map.Entry<Integer, Integer>> list = new ArrayList<>(map.entrySet());
    list.sort((o1, o2) -> !o1.getValue().equals(o2.getValue()) ? Integer.compare(o2.getValue(), o1.getValue()) : Integer.compare(o1.getKey(), o2.getKey()));
    Map<Integer, Integer> newMap = new LinkedHashMap<>();
    for (Map.Entry<Integer, Integer> entry : list)
        newMap.put(entry.getKey(), entry.getValue());
    return newMap;
}