[25]K 个一组翻转链表

# 给你一个链表，每 k 个节点一组进行翻转，请你返回翻转后的链表。 
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#  k 是一个正整数，它的值小于或等于链表的长度。 
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#  如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。 
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#  进阶： 
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#  你可以设计一个只使用常数额外空间的算法来解决此问题吗？ 
#  你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。 
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#  示例 1： 
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# 输入：head = [1,2,3,4,5], k = 2
# 输出：[2,1,4,3,5]
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#  示例 2： 
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# 输入：head = [1,2,3,4,5], k = 3
# 输出：[3,2,1,4,5]
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#  示例 3： 
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# 输入：head = [1,2,3,4,5], k = 1
# 输出：[1,2,3,4,5]
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#  示例 4： 
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# 输入：head = [1], k = 1
# 输出：[1]
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#  提示： 
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#  
#  列表中节点的数量在范围 sz 内 
#  1 <= sz <= 5000 
#  0 <= Node.val <= 1000 
#  1 <= k <= sz 
#  
#  Related Topics 递归 链表 👍 1334 👎 0


# leetcode submit region begin(Prohibit modification and deletion)
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        # 拆分 pointer,...,head,head.next,...
        count = 1
        pointer = head
        while count < k:
            head = head.next
            if head is None:
                return pointer
            count += 1
        next_head = head.next
        if next_head is None:
            return self.reverseList(pointer)
        # 断开
        head.next = None
        # 翻转
        reversed_head = self.reverseList(pointer)
        # 连接
        pointer.next = self.reverseKGroup(next_head, k)
        return reversed_head

    def reverseList(self, head) -> ListNode:
        # 翻转链表
        if head is None:
            return head
        if head.next is None:
            return head
        new_head = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return new_head
# leetcode submit region end(Prohibit modification and deletion)