实现一个链表，最多存储 10 个名称，先入先出。然后实现两种方法，其中一种方法是按姓氏的字母顺序对其进行排序。这是我遇到麻烦的地方。这是我尝试过的：
递归。该方法调用两个节点，比较它们，如果需要交换，然后调用自身。不适用于奇数个名称，并且往往是完整的错误。
收藏，但我对它的了解还不够，无法有效地使用它。
排序算法（例如冒泡排序）：我可以浏览列表，但很难让节点交换。
我的问题是：最简单的方法是什么？
public class List
{
public class Link
{
public String firstName;
public String middleName;
public String lastName;
public Link next = null;
Link(String f, String m, String l)
{
firstName = f;
middleName = m;
lastName = l;
}
}
private Link first_;
private Link last_;
List()
{
first_ = null;
last_ = null;
}
public boolean isEmpty()
{
return first_ == null;
}
public void insertFront(String f, String m, String l)
{
Link name = new Link(f, m, l);
if (first_ == null)
{
first_ = name;
last_ = name;
}
else
{
last_.next = name;
last_ = last_.next;
}
}
public String removeFront()
{
String f = first_.firstName;
String m = first_.middleName;
String l = first_.lastName;
first_ = first_.next;
return f + " " + m + " " + l;
}
public String findMiddle(String f, String l)
{
Link current = first_;
while (current != null && current.firstName.compareTo(f) != 0 && current.lastName.compareTo(l) != 0)
{
current = current.next;
}
if (current == null)
{
return "Not in list";
}
return "That person's middle name is " + current.middleName;
}
}
public class NamesOfFriends
{
public static void main(String[] args)
{
List listOfnames = new List();
Scanner in = new Scanner(System.in);
for(int i = 0; i < 3; i++)
{
if(i == 0)
{
System.out.println("Please enter the first, middle and last name?");
listOfnames.insertFront(in.next(), in.next(),in.next());
}
else
{
System.out.println("Please enter the next first, middle and last name");
listOfnames.insertFront(in.next(), in.next(),in.next());
}
}
System.out.println("To find the middle name, please enter the first and last name of the person.");
System.out.println(listOfnames.findMiddle(in.next(),in.next()));
}
}
编辑
经过一番努力，我想出了如何对它进行排序。为此，我正在尝试实现一种删除方法，该方法可以删除列表中任何位置的节点。虽然它确实编译，但当我运行程序时它什么也不做。
public Link remove(String lastName)
{
Link current_ = first_;
Link prior_ = null;
Link temp_ = null;
while (current_ != null && current_.lastName.compareTo(lastName) != 0)
{
prior_ = current_;
current_ = current_.next;
}
if (current_ != null)
{
if (current_ == last_)
{
temp_ = last_;
last_ = prior_;
}
else if (prior_ == null)
{
temp_ = first_;
first_ = first_.next;
}
}
else
{
temp_ = current_;
prior_.next = current_.next;
}
return temp_;
}