符号表

Posted 2020-11-05 ppwq

无序链表的顺序查找

public class SequentialSearchST<Key, Value>
{
	private Node first;           //链表首结点
	private class Node
	{	//链表结点的定义
		Key key;
		Value val;
		Node next;
		public Node(Key key, Value val, Node next)
		{
			this.key = key;
			this.val = val;
			this.next = next;
		}
	}
	public Value get(Key key)
	{	//查找给定的键，返回相关联的值
		for (Node x = first; x != null; x = x.next)
			if (key.equals(x.key))
				return x.val;        //命中
		return null;                 //未命中
	}
	public void put(key key, Value val)
	{	//查找给定的键，找到则更新其值，否则在表中新建结点
		for (Node x = first; x != null; x = x.next)
			if (key.equals(x.key))
			{	x.val = val; return;	}     //命中，更新
		first = new Node(key, val, first);    //未命中，新建结点
	}
}

　　向一个空表插入N个不同的键需要N²/2次比较，一次查找所需比较数，采用随机命中的话是N/2，说明基于链表的实现和顺序查找是非常低效的。

有序数组中的二分查找

 

public class BinarySearchST<Key extends Comparable<Key>, Value>
{
	private Key[] keys;
	private Value[] vals;
	private int N;
	public BinarySearchST(int capacity)
	{
		keys = (Key[])  new Comparable[capacity];
		vals = (value[]) new Object[capacity];
	}
	public int size()
	{	return N;	}
	public Value get(Key key)
	{
		if (isEmpty()) return null;
		int i = rank(key);
		if (i < N && keys[i].compareTo(key) == 0) return vals[i];
		else                                      return null;
	}
	public int rank(Key key)
	{
		int lo = 0, hi = N-1;
		while (lo <= hi)
		{
			int mid = lo + (hi - lo) / 2;
			int cmp = key.compareTo(keys[mid]);
			if      (cmp < 0) hi = mid - 1;
			else if (cmp > 0) lo = mid + 1;
			else return mid;
		}
		return lo;
	}
	public void put(Key key, Value value)
	{	//查找键，找到则更新值，否则创建新的元素
		int i = rank(key);
		if (i < N && keys[i].compareTo(key) == 0)
		{	vals[i] = val; return;	}
		for (int j = N; j > i; j--)
		{	keys[j] = keys[j-1]; vals[j] = vals[j-1];	}
		keys[i] = key; vals[i] = val;
		N++;
	}
	public Key min()
	{	return keys[0];	}
	public Key max()
	{	return keys[N-1];	}
	public Key select(int k)
	{	return keys[k];	}
	public Key ceiling(Key key)
	{
		int i = rank(key);
		return keys[i];
	}
	public Key floor(Key key)
	{
	}
	public void delete(Key key)
	{
		int i = rank(key);
		for (int j = i; j < N-1; j++)
		{	keys[j] = keys[j+1]; vals[j] = vals[j+1];	}
		
	}
	public Iterable<Key> keys(Key lo, Key hi)
	{
		Queue<Key> q = new Queue<Key>();
		for (int i = rank(lo); i < rank(hi); i++)
			q.enqueue(keys[i]);
		if (contains(hi))
			q.enqueue(keys[rank(hi)]);
		return q;
	}
}

　　N个键的有序数组进行二分查找最多需要（lgN+1）次比较。