给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中 没有重复出现 的数字。
示例 1:
输入: 1->2->3->3->4->4->5
输出: 1->2->5
示例 2:
输入: 1->1->1->2->3
输出: 2->3
在真实的面试中遇到过
/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode(int x) { val = x; }* }*/思路：递归
class Solution {int val = -100;public ListNode deleteDuplicates(ListNode head) {if (head == null) return null;ListNode node = deleteDuplicates(head.next);head.next = null;if (node == null){if (head.val == val) {return null;} else return head;}if (node.val == head.val) {val = head.val;return node.next;}if (head.val == val) {val = head.val;return node;}head.next = node;return head;}
}
2.递归
public ListNode deleteDuplicates(ListNode head) {if (head == null || head.next == null) return head;if (head.val == head.next.val) {ListNode nenode = head.next;while (nenode.next != null && nenode.val == nenode.next.val) {nenode = nenode.next;} return deleteDuplicates(nenode.next); } else {head.next = deleteDuplicates(head.next);}return head;}
迭代
public ListNode deleteDuplicates(ListNode head) {if (head == null || head.next == null) return head;ListNode root = new ListNode(-100);root.next = head;ListNode cur = head;ListNode pre = root;while (cur != null) {if (cur.next != null && cur.val == cur.next.val) {while (cur.next != null && cur.val == cur.next.val) {cur = cur.next;}pre.next = cur.next;cur = cur.next;} else {cur = cur.next;pre = pre.next;}}return root.next;}