【试题描述】定义一个函数，输入一个链表，删除无序链表中重复的节点
【参考代码】
方法一：
Without a buffer, we can iterate with two pointers: “current” does a normal iteration, while
“runner” iterates through all prior nodes to check for dups Runner will only see one dup
per node, because if there were multiple duplicates they would have been removed already
1 public static void deleteDups(LinkList head)
2 {
3 if (head == null)
4 return;
5 Link previous = head.first;
6 Link current = previous.next;
7 while (current != null)
8 {
9 Link runner = head.first;
10 while (runner != current)
11 {
12 if (runner.id == current.id)
13 {
14 Link tmp = current.next;
15 previous.next = tmp;
16 current = tmp;
17 break;
18 }
19 runner = runner.next;
20 }
21
22 if (runner == current)
23 {
24 previous = current;
25 current = current.next;
26 }
27 }
28
29 System.out.println("-----------");
30 head.displayList();
31 }
方法二：
If we can use a buffer, we can keep track of elements in a hashtable and remove any dups:
public static void deleteDups2(LinkList head)
{
if (head == null)
return;
Hashtable table = new Hashtable();
Link previous = null;
Link current = head.first;
while (current != null)
{
if (table.containsKey(current.id))
previous.next = current.next;
else
{
table.put(current.id, true);
previous = current;
}
current = current.next;
}
System.out.println("-----------");
head.displayList();
}