0

我试图找到一种方法来按已输入的姓氏对这个单链表进行排序。我想我可能会尝试冒泡排序，但我在通过第二个元素遍历列表并进行比较时遇到问题。该列表现在仅包含 3 个名称作为控制台条目，但完成后它应该有 10 个名称。任何帮助将不胜感激。

package LinkedList;

import java.util.*;


class SLinkedList
{

   public String data1;
   public String data2;
   public SLinkedList next;

   public SLinkedList()
   {
      data1 = "";
      data2 = "";
      next = null;
   }
   public SLinkedList(String value1, String value2)
   {
      data1 = value1;
      data2 = value2;
      next = null;
   }
   public SLinkedList InsertNext(String value1, String value2)
   {
      SLinkedList node = new SLinkedList(value1, value2);
      if(this.next == null)
      {
         // Easy to handle
         node.next = null; // already set in constructor
         this.next = node;
      }
      else
      {
         // Insert in the middle
         SLinkedList temp = this.next;
         node.next = temp;
         this.next = node;
       }
       return node;
   }


   public int DeleteNext()
   {
      if(next == null)
         return 0;
       SLinkedList node = this.next;
       this.next = this.next.next;  // can be NULL here
       node = null;
       return 1;
   }
   public void Traverse(SLinkedList node)
   {
      if(node == null)
         node = this;
      System.out.println("\nTraversing in Forward Direction\n");
      while(node != null)
      {
         System.out.println(node.data1 + " " + node.data2);
         node = node.next;
      }
   }
   public void bubbleSort(SLinkedList node) {
       if(node == null)
           node = this;
       String current;
       String second;
       String temp;
       System.out.println("Attemptint to sort...");
       while(node != null)
       {
           current = node.data2;
           node = node.next;
           second = node.data2;
           System.out.println(current + " " + second);
           if(current.compareTo(second) < 0) {

               System.out.println("greater than zero");
           }

           node = null;
           //node = node.next;

       }    
   }


      public static void main(String[] args)
   {

      String firstName;
      String lastName;
      Scanner sc = new Scanner(System.in);
      System.out.print("Enter names in the format of: 'Ed King' with"
              + " a single space in between.");
      firstName = sc.next();
      lastName = sc.next();
      SLinkedList node1 = new SLinkedList(firstName, lastName);
      System.out.print("Enter second name: ");
      firstName = sc.next();
      lastName = sc.next();
      SLinkedList node2 = node1.InsertNext(firstName, lastName);
      System.out.print("Enter third name: ");
      firstName = sc.next();
      lastName = sc.next();
      SLinkedList node3 = node2.InsertNext(firstName, lastName);


      node1.bubbleSort(null);

   }
}
4

2 回答 2

0

首先应该使用归并排序对链表进行排序。所以改变你的算法。

来到冒泡排序实现，你的实现不正确[为什么你根本不使用 temp ]。您应该寻找交换节点。

于 2013-03-31T13:57:53.657 回答
0

提示：

  • 您对冒泡排序的实现从根本上是错误的。冒泡排序需要一个嵌套循环。

  • 鉴于您设计数据结构的方式，冒泡排序方法必须返回列表的新开始。

  • 不要以大写字母开头的方法名称。这是 Java 而不是 C#。

  • 如果您无法可视化代码在做什么，请使用调试器单步执行并观察状态如何变化。

于 2013-03-31T04:23:18.507 回答