Leetcode----- 82.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.

给定一个排序的链表，删除所有具有重复数字的节点，从原始列表中只留下不同的结点。

代码如下:

public class RemoveDuplicatesfromSortedListII {    public static ListNode deleteDuplicates(ListNode head) {    if(head == null || head.next == null) {    return head;    }    ListNode ret = new ListNode(0);    ListNode node = ret;    ListNode pre = head;    ListNode point = head.next;    while(point != null) {    if(pre.val == point.val) {    while(point.next != null && point.next.val == pre.val) {    point = point.next;    }pre = point.next;ret.next = pre;point = point.next;    }else {    ret.next = pre;    pre = pre.next;    ret = ret.next;    }    if(point != null) {    point = point.next;    }     }        return node.next;    }public static void main(String[] args) {ListNode l10 = new ListNode(1);ListNode l11 = new ListNode(1);ListNode l12 = new ListNode(1);ListNode l13 = new ListNode(1);ListNode l14 = new ListNode(1);ListNode l15 = new ListNode(1);ListNode l16 = new ListNode(1);l10.next = l11;l11.next = l12;l12.next = l13;l13.next = l14;l14.next = l15;l15.next = l16;l16.next = null;ListNode node = deleteDuplicates(l10);while(node != null) {if(node.next == null) {System.out.println(node.val);}else{System.out.print(node.val +"->");}node = node.next;}}}