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。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ListNode header = new ListNode(-1);
        header.next = null;
        ListNode cur = header, per = head;
        while(per != null){
            if(per.next == null || per.val != per.next.val){
                cur.next = per;
                cur = cur.next;
                per = per.next;
                cur.next = null;
            }else{
                int value = per.val;
                while(per != null && per.val == value){
                    per = per.next;
                }
            }
        }
        return header.next;
    }
}