Remove Duplicates from Sorted List II

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct 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.

Solution: 指针操作。

如果当前val和下一个val相等，则将per指针一直移到和这些val不等的位置。

如果当前val和下一个不等， 则这个val要保存。 那么 先将cur.next = per;  cur = cur.next; per = per.next; 注意还要让cur.next = null;这样才保证了不会加上些不需要的元素，同时 这个句子只能放在最后。

 1 /**

 2  * Definition for singly-linked list.

 3  * public class ListNode {

 4  *     int val;

 5  *     ListNode next;

 6  *     ListNode(int x) {

 7  *         val = x;

 8  *         next = null;

 9  *     }

10  * }

11  */

12 public class Solution {

13     public ListNode deleteDuplicates(ListNode head) {

14         // IMPORTANT: Please reset any member data you declared, as

15         // the same Solution instance will be reused for each test case.

16         ListNode header = new ListNode(-1);

17         header.next = null;

18         ListNode cur = header, per = head;

19         while(per != null){

20             if(per.next == null || per.val != per.next.val){

21                 cur.next = per;

22                 cur = cur.next;

23                 per = per.next;

24                 cur.next = null;

25             }else{

26                 int value = per.val;

27                 while(per != null && per.val == value){

28                     per = per.next;

29                 }

30             }

31         }

32         return header.next;

33     }

34 }