2022-02-26k个一组反转链表

可以借鉴以前的反转链表的代码，然后分别一组一组的反转。
属于细节题目，适合反复训练

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head, tail):
        if not head:
            return head
        # 注意这里需要保留与tail之后数据的连接
        pre = tail.next
        p = head
        while pre != tail:
            q = p.next
            p.next = pre
            pre = p
            p = q
        return pre, head

    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        dummy = ListNode()
        dummy.next = head
        left = dummy
        tail = dummy
        # print('left,right,head',left.val, right.val,head.val)
        while head:
            for i in range(k):
                tail = tail.next
                if not tail:
                    return dummy.next
            # print('head,tail',head,tail)
            # 开始翻转前先记录被翻转部分的左右两部分
            right = tail.next
            new_head, new_tail = self.reverseList(head, tail)
            # print('new_head,new_tail',new_head,new_tail)
            # 翻转后的新链表连接到原链表中间
            left.next = new_head
            new_tail.next = right

            # 重新初始化待翻转链表的head、tail以及左半部分
            head = new_tail.next
            tail = new_tail

            left = new_tail
        return dummy.next