算法－第四版－练习1.3.24解答

问题

编写一个方法removeAfter()，接受一个链表结点作为参数并删除该结点的后续结点（如果参数结点或参数结点的后续结点为空则什么也不做）。

解决思路

迭代删除其后续结点。

代码

    public void removeAfter(Node<Item> node)
    {
        if (node == null || node.next == null)
            return;
        Node<Item> current = node.next;
        Node<Item> next = current.next;
        node.next = null;
        while (current.next != null)
        {
            current = null;
            current = next;
            next = next.next;
        }
    }
    
    public Node<Item> search(Item item)
    {
        Node<Item> current = first;
        while (current != null)
        {
            if (item.equals(current.item))
            {
                return current;
            }
            current = current.next;
        }
        return null;
    }

测试代码：

/**
 * Description : 
 * Author      : mn@furzoom.com
 * Date        : Oct 24, 2016 6:06:52 PM
 * Copyright (c) 2013-2016, http://furzoom.com All Rights Reserved.
 */
package com.furzoom.lab.algs.ch103;

/**
 * ClassName    : E10324 <br>
 * Function     : TODO ADD FUNCTION. <br>
 * date         : Oct 24, 2016 6:06:52 PM <br>
 * 
 * @version 
 */
public class E10324
{
    public static void main(String[] args)
    {
        LinkList<String> ll = new LinkList<String>();
        ll.append("a");
        ll.append("B");
        ll.append("c");
        ll.append("D");
        ll.append("e");
        
        LinkList.printList(ll);
        
        Node<String> node = ll.search("f");
        ll.removeAfter(node);
        System.out.println("remove after \"f\": ");
        LinkList.printList(ll);
        
        node = ll.search("e");
        ll.removeAfter(node);
        System.out.println("remove after \"e\": ");
        LinkList.printList(ll);
        
        node = ll.search("D");
        ll.removeAfter(node);
        System.out.println("remove after \"D\": ");
        LinkList.printList(ll);
        
        node = ll.search("a");
        ll.removeAfter(node);
        System.out.println("remove after \"a\": ");
        LinkList.printList(ll);
    }
           
}

输出：

a
B
c
D
e
remove after "f": 
a
B
c
D
e
remove after "e": 
a
B
c
D
e
remove after "D": 
a
B
c
D
remove after "a": 
a

算法－第四版－1.3 背包、队列和栈－习题索引汇总

算法－第四版习题索引汇总