README_CN.md

25. K 个一组翻转链表

给你链表的头节点 head ，每 k 个节点一组进行翻转，请你返回修改后的链表。

k 是一个正整数，它的值小于或等于链表的长度。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。

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。

示例 1:

输入: head = [1,2,3,4,5], k = 2
输出: [2,1,4,3,5]

示例 2:

输入: head = [1,2,3,4,5], k = 3
输出: [3,2,1,4,5]

提示:

• 链表中的节点数目为 n
• 1 <= k <= n <= 5000
• 0 <= Node.val <= 1000

**进阶：**你可以设计一个只用 O(1) 额外内存空间的算法解决此问题吗？

题解 (Python)

1. 题解

# 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: Optional[ListNode], k: int) -> Optional[ListNode]:
        length = 0
        curr = head
        while curr is not None:
            length += 1
            curr = curr.next

        dummy = ListNode(next=head)
        grouptail = dummy

        for _ in range(length // k):
            grouphead = grouptail.next
            prev = grouphead
            for _ in range(k):
                prev = prev.next

            curr = grouphead
            for _ in range(k):
                temp = curr
                curr = curr.next
                temp.next = prev
                prev = temp

            grouptail.next = prev
            grouptail = grouphead

        return dummy.next