solution.md

# 删除排序链表中的重复元素 II

<p>存在一个按升序排列的链表，给你这个链表的头节点 <code>head</code> ，请你删除链表中所有存在数字重复情况的节点，只保留原始链表中 <strong>没有重复出现</strong><em> </em>的数字。</p><p>返回同样按升序排列的结果链表。</p><p> </p><p><strong>示例 1：</strong></p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0082.Remove%20Duplicates%20from%20Sorted%20List%20II/images/linkedlist1.jpg" style="width: 500px; height: 142px;" /><pre><strong>输入：</strong>head = [1,2,3,3,4,4,5]<strong><br />输出：</strong>[1,2,5]</pre><p><strong>示例 2：</strong></p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0082.Remove%20Duplicates%20from%20Sorted%20List%20II/images/linkedlist2.jpg" style="width: 500px; height: 205px;" /><pre><strong>输入：</strong>head = [1,1,1,2,3]<strong><br />输出：</strong>[2,3]</pre><p> </p><p><strong>提示：</strong></p><ul>	<li>链表中节点数目在范围 <code>[0, 300]</code> 内</li>	<li><code>-100 <= Node.val <= 100</code></li>	<li>题目数据保证链表已经按升序排列</li></ul>

## template

```java
public class ListNode {
	int val;
	ListNode next;
	ListNode(int x) { val = x; }
}
class Solution {
	public ListNode deleteDuplicates(ListNode head) {
		if (head == null || head.next == null) {
			return head;
		}
		ListNode next = head.next;
		if (head.val == next.val) {
			while (next != null && head.val == next.val) {
				next = next.next;
			}
			head = deleteDuplicates(next);
		} else {
			head.next = deleteDuplicates(next);
		}
		return head;
	}
}
```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```