# 剑指offer18：删除链表节点

![在这里插入图片描述](https://img-blog.csdnimg.cn/20190718165927375.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2phY2tpZV9vMm8y,size_16,color_FFFFFF,t_70) 1.考虑输入空链表和一个节点链表 2.如果头节点不重复，直接递归查找重复 3.双指针，进行判断两个节点是不是相等

```python
# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def deleteDuplication(self, pHead):
        # write code here
        if pHead is None or pHead.next is None:
            return pHead
        head1 = pHead.next
        if head1.val != pHead.val:
            pHead.next = self.deleteDuplication(pHead.next)
        else:
            while pHead.val == head1.val and head1.next is not None:
                head1 = head1.next
            if head1.val != pHead.val:
                pHead = self.deleteDuplication(head1)
            else:
                return None
        return pHead
```


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