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Write code to remove duplicates from an unsorted linked list
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How would you solve this problem if a temporary buffer is not allowed?
题解如下：
解决此题关键是了解java reference及链表基本知识
链表类：
public class LinkedListNode { LinkedListNode next=null; int data; public LinkedListNode(int d) { data=d; } public void appendToTail(int d) { LinkedListNode end= new LinkedListNode(d); LinkedListNode node=this; while(node.next!=null) { node=node.next; } node.next=end; } public static void printList(LinkedListNode node) { LinkedListNode tempnode=node; } public static LinkedListNode deleteNode(LinkedListNode head,int d) { LinkedListNode node=head; if(node.data==d) return node.next; while(node.next!=null) { if(node.next.data==d) { node.next=node.next.next; return head; } node=node.next; } return null; } }
问题解法：
import java.util.HashSet; public class RemoveDupsFromList { public static void deleteDups(LinkedListNode n) { HashSet<Integer>set=new HashSet<>(); LinkedListNode previous=null; while (n!=null) { if(set.contains(n.data)) previous.next=n.next; else { set.add(n.data); previous=n; } n=n.next; } } //without buffer public static void deleteDups2(LinkedListNode n) { if(n==null) return; LinkedListNode head,pre,cur,tmp; head=pre=cur=n; while (cur!=null) { tmp=head; while (tmp!=cur) { if(cur.data==tmp.data) { pre.next=cur.next; cur=cur.next; break; } tmp=tmp.next; } if(tmp==cur) { pre=cur; cur=cur.next; } } } public static void deleteDups3(LinkedListNode n) { } public static void main(String[] args) { LinkedListNode node=new LinkedListNode(5); node.appendToTail(4); node.appendToTail(3); node.appendToTail(3); node.appendToTail(4); node.appendToTail(2); node.appendToTail(2); LinkedListNode node1=node; deleteDups2(node); System.out.println(node.data); } public static void test(LinkedListNode node) { node=node.next; System.out.println(node.data); } }
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原文地址：http://www.cnblogs.com/AlexWei/p/4946219.html
