哈希表（2）——哈希表动态空间处理和复杂度分析

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/u012292754/article/details/87189329

1 哈希表复杂度分析（链地址法）

一共有 M 个地址，如果放入的总元素个数是 N

• 如果每个地址是链表，O(N / M)
• 如果每个地址是平衡树：O(log( N / M))
  

2 哈希表的动态空间处理

• 当平均每个地址承载的元素超过一定程度，即扩容，N / M >= upperTol
• 当平均每个地址承载的元素少过一定程度，即缩容， N / M < lowerTol
• HashTable.java

package hashtable;

import java.util.TreeMap;

public class HashTable<K, V> {

    private static final int upperTol = 10;
    private static final int lowerTol = 2;
    private static final int initCapacity = 7;

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

    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(initCapacity);
    }

    private int hash(K key) {
        // 负数可以变成正数
        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++;

            if (size >= upperTol * M) {
                resize(2 * M);
            }
        }
    }

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

            if (size < lowerTol * M && M / 2 >= initCapacity) {
                resize(M / 2);
            }

        }

        return ret;
    }

    public void set(K key, V value) {
        TreeMap<K, V> map = hashtable[hash(key)];

        if (!map.containsKey(key)) {
            throw new IllegalArgumentException(key + "not 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);
    }

    private void resize(int newM) {

        TreeMap<K, V>[] newHashTable = new TreeMap[newM];

        for (int i = 0; i < newM; i++) {
            newHashTable[i] = new TreeMap<>();
        }

        int oldM = M;
        this.M = newM;
        for (int i = 0; i < oldM; i++) {

            TreeMap<K, V> map = hashtable[i];

            for (K key : map.keySet()) {
                newHashTable[hash(key)].put(key, map.get(key));
            }
        }

        this.hashtable = newHashTable;
    }


}

3 更复杂的动态空间处理

• 2 * M 不是素数
• 解决方式：按照下表扩容
  
• HashTable.java

package hashtable;

import java.util.TreeMap;

public class HashTable<K, V> {


    private final int[] capacity = {
            53, 97, 193, 389, 769, 1543, 3079, 6151, 12289, 24593,
            49157, 98317, 196613, 393241, 786433, 1572869, 3145739, 6291469,
            12582917, 25165843, 50331653, 100663319, 201326611, 402653189,
            805306457, 1610612741};

    private static final int upperTol = 10;
    private static final int lowerTol = 2;
    private int capacityIndex = 0;

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

    public HashTable() {
        this.M = capacity[capacityIndex];
        size = 0;
        hashtable = new TreeMap[M];

        for (int i = 0; i < M; i++) {
            hashtable[i] = new TreeMap<>();
        }
    }


    private int hash(K key) {
        // 负数可以变成正数
        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++;

            if (size >= upperTol * M && capacityIndex + 1 < capacity.length) {
                capacityIndex++;
                resize(capacity[capacityIndex];
            }
        }
    }

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

            if (size < lowerTol * M && capacityIndex - 1 >= 0) {
                capacityIndex--;
                resize(capacity[capacityIndex]);
            }

        }

        return ret;
    }

    public void set(K key, V value) {
        TreeMap<K, V> map = hashtable[hash(key)];

        if (!map.containsKey(key)) {
            throw new IllegalArgumentException(key + "not 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);
    }

    private void resize(int newM) {

        TreeMap<K, V>[] newHashTable = new TreeMap[newM];

        for (int i = 0; i < newM; i++) {
            newHashTable[i] = new TreeMap<>();
        }

        int oldM = M;
        this.M = newM;
        for (int i = 0; i < oldM; i++) {

            TreeMap<K, V> map = hashtable[i];

            for (K key : map.keySet()) {
                newHashTable[hash(key)].put(key, map.get(key));
            }
        }

        this.hashtable = newHashTable;
    }


}

4 哈希表

• 均摊复杂度为 O(1)
• 牺牲了 顺序性
  

5 自定义的哈希表的 bug

5.1 Java8 中的哈希表

• Java8 之前每一个位置对应一个链表；Java8 开始，当哈希冲突达到一定的程度，每一个位置从链表转成红黑树；
• 因为初始的是链表，所以不要求 K 是 Comparable;
• 转成红黑树的条件：K 是 Comparable,否则依然保持链表；