求最高频的kth个字符串：利用PriorityQueue和 HashMap

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本文介绍了一种使用PriorityQueue和HashMap实现的高效算法，用于找出列表中出现频率最高的K个单词。通过具体代码示例，展示了如何统计单词频率并获取最高频的单词，适用于web日志分析等场景。

LintCode 471. 最高频的K个单词

题目：给一个单词列表，求出这个列表中出现频次最高的K个单词。

问题：web 日志，求访问频次最高的kth个URI和频次。

求kth个频次最高的字符串：利用PriorityQueue和 HashMap

时间复杂度 nlogn, 利用PQ二叉堆排序好于快速排序。

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public class PriorityQueueTest {
    public static void main(String[] args) {

        int m=20;
        int k=3;
        //int[] a = {3,3,1,2,5,6,7,8,9,9,11,11,11,12,13,14,19,16,17,20};//m个元素
        String[] s={
                "hello","hello","hello","world","the",
                "You", "should", "not","use","the",
                "Know", "someone", "You","can", "answer",
                "Share","link", "to", "this","email" };

        URIQueue myQQ =new URIQueue(s,m);
        myQQ.getHotURI(k);
    }
}

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import java.util.Comparator;
import java.util.HashMap;
import java.util.PriorityQueue;

//统计URI 频次高的项!
public class URIQueue {

    private HashMap<String, Integer> rawURI;
    private PriorityQueue<String> pq;
    private int mItem;

    public URIQueue(String[] s, int mm) {

        System.out.println("\nURIQueue\n");

        mItem = mm;
        rawURI = new HashMap<>(mItem);

        for (int i = 0; i < mItem; i++) {
            int count = 1;
            if (rawURI.containsKey(s[i])) {
                count = rawURI.get(s[i]) + 1;
            }
            rawURI.put(s[i], count);
        }
        System.out.println(rawURI);

        URIComparator<String> strCompare = new URIComparator<>();
        pq = new PriorityQueue<>(mItem, strCompare); //Line x
    }

    public void getHotURI(int kth) {

        for (String s1 : rawURI.keySet()) {
            pq.add(s1);
        }
        System.out.println(pq);

        int count=0;
        while (!pq.isEmpty() && count<kth) {
            count++;
            String sKey = pq.remove();
            System.out.printf("#%d:%s=%d\n",count,sKey,rawURI.get(sKey));
        }
    }


    public class URIComparator<String> implements Comparator<String> {

        public int compare(String key1, String key2) {
            if (rawURI.get(key2) - rawURI.get(key1) > 0) {
                return 1;
            } else if (rawURI.get(key2) - rawURI.get(key1) < 0) {
                return -1;
            }
            return 0;
        }
    }
}

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运行结果：

URIQueue

{use=1, Know=1, link=1, this=1, the=2, can=1, not=1, world=1, someone=1, answer=1, should=1, hello=3, to=1, Share=1, You=2, email=1}
[hello, use, the, this, Know, link, You, world, someone, answer, should, can, to, Share, not, email]
#1:hello=3
#2:the=2
#3:You=2

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其他解法：

https://blog.youkuaiyun.com/qq_27139155/article/details/79754671

利用Collections.sort ，平均时间复杂度 O(nlogn)