TopK问题

实现思路：

1. 选择小顶堆：维护当前最高的K个元素，堆顶是这K个中的最小值。新元素若大于堆顶，则替换堆顶并调整堆，确保堆中始终为Top K大元素。

2. 插入逻辑：堆未满时直接插入；堆满时，新元素大于堆顶则替换，否则忽略。

3. 获取排行榜：取出堆元素并按热度降序排列。

import java.util.*;

public class HotSearchRanking {

    static class HotSearchItem {
        private final String keyword;
        private final int heat;

        public HotSearchItem(String keyword, int heat) {
            this.keyword = keyword;
            this.heat = heat;
        }

        public int getHeat() {
            return heat;
        }

        public String getKeyword() {
            return keyword;
        }

        @Override
        public String toString() {
            return keyword + "(" + heat + ")";
        }
    }

    private final PriorityQueue<HotSearchItem> minHeap;
    private final int k;

    public HotSearchRanking(int k) {
        this.k = k;
        // 初始化小顶堆，按热度升序排序
        this.minHeap = new PriorityQueue<>(Comparator.comparingInt(HotSearchItem::getHeat));
    }

    public void add(HotSearchItem item) {
        if (item == null) return;
        if (minHeap.size() < k) {
            minHeap.offer(item);
        } else {
            // 仅当新元素热度大于堆顶时替换
            if (item.getHeat() > minHeap.peek().getHeat()) {
                minHeap.poll();
                minHeap.offer(item);
            }
        }
    }

    public List<HotSearchItem> getTopK() {
        List<HotSearchItem> result = new ArrayList<>(minHeap);
        // 按热度降序排序以形成排行榜
        result.sort(Comparator.comparingInt(HotSearchItem::getHeat).reversed());
        return result;
    }

    public static void main(String[] args) {
        HotSearchRanking ranking = new HotSearchRanking(3);
        ranking.add(new HotSearchItem("Java", 50));
        ranking.add(new HotSearchItem("Python", 30));
        ranking.add(new HotSearchItem("C++", 70));
        ranking.add(new HotSearchItem("JavaScript", 60));
        ranking.add(new HotSearchItem("Go", 65));

        List<HotSearchItem> top = ranking.getTopK();
        System.out.println("实时热搜排行榜TOP3:");
        for (HotSearchItem item : top) {
            System.out.println(item);
        }
    }
}

该方法确保每次插入和调整的时间复杂度为O(log K)，获取排行榜的时间复杂度为O(K log K)，适合处理实时更新的热搜数据。