算法－第四版－练习1.3.30解答

问题

编写一个函数，接受一条链表的首结点作为参数，（破坏性地）将链表反转并返回结果链表的首结点。

解决思路

代码

public Node<Item> reverse()
{
Node<Item> oldFirst = first;;
first = null;
while (oldFirst != null)
{
Node<Item> second = oldFirst.next;
oldFirst.next = first;
first = oldFirst;
oldFirst = second;
}
return first;
}


测试代码：
/**
* Description :
* Author : mn@furzoom.com
* Date : Oct 25, 2016 3:08:48 PM
* Copyright (c) 2013-2016, http://furzoom.com All Rights Reserved.
*/
package com.furzoom.lab.algs.ch103;

/**
* ClassName : E10330 <br>
* Function : TODO ADD FUNCTION. <br>
* date : Oct 25, 2016 3:08:48 PM <br>
*
* @version
*/
public class E10330
{
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");

ll.printList();
ll.reverse();
System.out.println("After reverse list:");
ll.printList();
}
}


结果：
a
B
c
D
e
After reverse list:
e
D
c
B
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