Remove Duplicates from Sorted List II

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving onlydistinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

思路：注意：一定要做一个前指针，用来跳过所有相同的数，不然根据memory model，第一个相同的数的指针是不能delete自己的。指针指向会出问题，用cur 和curnext相比较来进行移动。

思路2：都一样，只是写法不一样，自己慢慢推倒，都可以推出来。

/** * 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) {        if(head == null || head.next == null) return head;        ListNode dump = new ListNode(0);        dump.next = head;        ListNode prehead = dump;        ListNode pre = head;        while(pre != null && pre.next!=null) {            ListNode cur = pre.next;            if(cur.val == pre.val) {                while(cur!=null && cur.val == pre.val) {                    cur = cur.next;                }                prehead.next = cur;                pre = cur;            } else {                prehead = pre;                pre = pre.next;            }        }        return dump.next;    }}

思路1：

/** * 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) {        if(head == null || head.next == null) return head;        ListNode dump = new ListNode(0);        dump.next = head;        ListNode pre = dump;        while(pre.next!=null && pre.next.next!=null) {            ListNode cur = pre.next;            ListNode curnext = pre.next.next;            if(cur.val != curnext.val) {                pre = pre.next;            } else {                while(curnext!=null && cur.val == curnext.val){                    curnext = curnext.next;                }                pre.next = curnext;            }        }        return dump.next;    }}