实现思路

• 维护一个M个元素的数组，数组的每一格挂一棵红黑树；
• 一个元素进来，先根据key计算出其要挂载到哪棵红黑树中；
• 确定了要挂载的红黑树后，问题就变成了在红黑树中添加，删除等操作了；
• M越小，哈希冲突的概率就越大，所以M的大小很大程度的影响了哈希表的效率；

import java.util.TreeMap;

public class HashTable<K, V> {

    private TreeMap<K, V>[] hashtable;
    private int size;
    private int M;

    public HashTable(int M){
        this.M = M;
        size = 0;
        hashtable = new TreeMap[M];
        for(int i = 0 ; i < M ; i ++)
            hashtable[i] = new TreeMap<>();
    }

    public HashTable(){
        this(97);
    }

    private int hash(K key){
        return (key.hashCode() & 0x7fffffff) % M;
    }

    public int getSize(){
        return size;
    }

    public void add(K key, V value){
        TreeMap<K, V> map = hashtable[hash(key)];
        if(map.containsKey(key))
            map.put(key, value);
        else{
            map.put(key, value);
            size ++;
        }
    }

    public V remove(K key){
        V ret = null;
        TreeMap<K, V> map = hashtable[hash(key)];
        if(map.containsKey(key)){
            ret = map.remove(key);
            size --;
        }
        return ret;
    }

    public void set(K key, V value){
        TreeMap<K, V> map = hashtable[hash(key)];
        if(!map.containsKey(key))
            throw new IllegalArgumentException(key + " doesn't exist!");

        map.put(key, value);
    }

    public boolean contains(K key){
        return hashtable[hash(key)].containsKey(key);
    }

    public V get(K key){
        return hashtable[hash(key)].get(key);
    }
}

哈希表性能测试

import java.util.ArrayList;

public class Main {

    public static void main(String[] args) {
        System.out.println("Pride and Prejudice");
        ArrayList<String> words = new ArrayList<>();
        if(FileOperation.readFile("pride-and-prejudice.txt", words)) {
            System.out.println("Total words: " + words.size());

            // Test BST
            long startTime = System.nanoTime();

            BST<String, Integer> bst = new BST<>();
            for (String word : words) {
                if (bst.contains(word))
                    bst.set(word, bst.get(word) + 1);
                else
                    bst.add(word, 1);
            }

            for(String word: words)
                bst.contains(word);

            long endTime = System.nanoTime();
            double time = (endTime - startTime) / 1000000000.0;
            System.out.println("BST: " + time + " s");


            // Test AVL
            startTime = System.nanoTime();

            AVLTree<String, Integer> avl = new AVLTree<>();
            for (String word : words) {
                if (avl.contains(word))
                    avl.set(word, avl.get(word) + 1);
                else
                    avl.add(word, 1);
            }

            for(String word: words)
                avl.contains(word);

            endTime = System.nanoTime();
            time = (endTime - startTime) / 1000000000.0;
            System.out.println("AVL: " + time + " s");


            // Test RBTree
            startTime = System.nanoTime();

            RBTree<String, Integer> rbt = new RBTree<>();
            for (String word : words) {
                if (rbt.contains(word))
                    rbt.set(word, rbt.get(word) + 1);
                else
                    rbt.add(word, 1);
            }

            for(String word: words)
                rbt.contains(word);

            endTime = System.nanoTime();
            time = (endTime - startTime) / 1000000000.0;
            System.out.println("RBTree: " + time + " s");


            // Test HashTable
            startTime = System.nanoTime();

            // HashTable<String, Integer> ht = new HashTable<>();
            HashTable<String, Integer> ht = new HashTable<>(131071);
            for (String word : words) {
                if (ht.contains(word))
                    ht.set(word, ht.get(word) + 1);
                else
                    ht.add(word, 1);
            }

            for(String word: words)
                ht.contains(word);

            endTime = System.nanoTime();
            time = (endTime - startTime) / 1000000000.0;
            System.out.println("HashTable: " + time + " s");
        }

        System.out.println();
    }
    
}

输出：

• 哈希表要稍微有点优势；

Pride and Prejudice
Total words: 125901
BST: 0.1774385 s
AVL: 0.0993948 s
RBTree: 0.0992375 s
HashTable: 0.0919606 s