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.

与Remove Duplicates from Sorted List相比，这道题目要求删除所有重复的元素，这样头结点可能会被删除，我们这时同样需要创建一个辅助节点，但是这次不是直接用辅助节点helper记录头结点，而是将helper.next = head，因为head可能会被删除，因此我们要返回的节点要根据helper.next来确定。这样相当于在原有链表的头上多加了一个节点，让head = helper，从head的开始操作，比较head.next的值与head.next.next的值，如果相等，用一个while循环来删除所有重复的元素；如果不等让head后移。最后返回helper.next。代码如下：

/**
 * 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) return head;
        ListNode helper = new ListNode(0);
        helper.next = head;
        head = helper;
        while(head != null && head.next != null && head.next.next != null) {
            if(head.next.val == head.next.next.val) {
                int value = head.next.val;
                while(head.next != null && head.next.val == value) {
                    head.next = head.next.next;
                }
            } else {
                head = head.next;
            }
        }
        return helper.next;
    }
}