删除重复节点

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,

Given 

1->1->2

, return 

1->2

.

Given 

1->1->2->3->3

, return 

1->2->3

.

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思路：后面的与当前的相同删除后面的即可，但是这种方法只能删除连续重复的（即1，1，1这种1，2，1这种无法删除）

代码：

import java.util.ArrayList;
/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode node = head;
for(;node!=null;node=node.next){
while(node.next!=null&&node.val==node.next.val){
if(node.next==null)node=null;
node.next = node.next.next;
}
}
return head;
}
}

递归的(有木有很赞啊！！虽然不是我写的（* ——*）)

public ListNode deleteDuplicates(ListNode head) {
if(head == null || head.next == null)return head;
head.next = deleteDuplicates(head.next);
return head.val == head.next.val ? head.next : head;
}

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode pre = new ListNode(-1);//to avoid the head check
pre.next = head;
ListNode cur = pre;
//ListNode node = head;
while(cur.next!=null){
if(cur.next.val==cur.val)cur.next = cur.next.next;
else
cur = cur.next;
}
return head;
}
}