力扣——Reverse Nodes in k-Group（K 个一组翻转链表） python实现

题目描述：

中文：

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。

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。

示例 :

给定这个链表：1->2->3->4->5

当 k = 2 时，应当返回: 2->1->4->3->5

当 k = 3 时，应当返回: 3->2->1->4->5

说明 :

你的算法只能使用常数的额外空间。
你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

英文：

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        pre = ListNode(-1)
        tail = pre
        
        q = head
        while q is not None:
            #向后查找k个节点
            n = k
            p = q
            while p is not None and n > 0:
                p = p.next
                n -= 1
                
            #如果在查找k个节点的过程中遇到None，则说明
            #后面的节点不够k个节点则直接跳出即可
            if n > 0:
                tail.next = q
                break
            
            #将这K个节点以头插法插入
            end = q
            while q != p:
                t = q.next
                q.next = tail.next
                tail.next = q
                q = t
            tail = end
        return pre.next

题目来源：力扣