链表

[92. 反转链表 II](https://leetcode.cn/problems/reverse-linked-list-ii/description/?envType=study-plan-v2&envId=top-interview-150)
```python
class Solution:
    def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
        p0 = dummy = ListNode(next = head)
        for i in range(left - 1):
            p0 = p0.next
        
        pre = None
        cur = p0.next
        for _ in range(right - left + 1):
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt
        p0.next.next = cur
        p0.next = pre
        return dummy.next
```
[82. 删除排序链表中的重复元素 II](https://leetcode.cn/problems/remove-duplicates-from-sorted-list-ii/?envType=study-plan-v2&envId=top-interview-150)
```python
class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        cur = dummy = ListNode(next=head)
        while cur.next and cur.next.next:
            val = cur.next.val
            if cur.next.next.val == val:
                while cur.next and cur.next.val == val:
                    cur.next = cur.next.next
            else:
                cur = cur.next
        return dummy.next
```
[61. 旋转链表](https://leetcode.cn/problems/rotate-list/?envType=study-plan-v2&envId=top-interview-150)
```python
class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        if not head or not head.next or k == 0:
            return head
        slow = fast = dummy = ListNode(next=head)
        cur = head
        n = 0
        while cur:
            n += 1
            cur = cur.next
        k = k % n if n != 0 else 0
        for _ in range(k):
            fast = fast.next
        while fast and fast.next:
            slow = slow.next
            fast = fast.next
        fast.next = head
        new_head = slow.next
        slow.next = None
        return new_head
```