k组倒置链表节点

Reverse Nodes in k-Group

• 给定一个链表，每k个倒置其节点，返回新链表的头部，不允许使用额外的内存，只能在原有链表节点基础上进行操作。

题目原文:

• 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.
• You may not alter the values in the nodes, only nodes itself may be changed.
• Only constant memory is allowed.

example 1

Given this linked list: 1->2->3->4->5For k = 2, you should return: 2->1->4->3->5For k = 3, you should return: 3->2->1->4->5

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思路

1. 参照我的博文成对交换链表节点，讨巧的办法是只交换节点内的value，不做节点的移动。
2. 如不使用上述方法，可以使用递归完成

代码

# Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = Noneclass Solution(object):    def reverseKGroup(self, head, k):        """        :type head: ListNode        :type k: int        :rtype: ListNode        """        try:            arr = []            tmp = head            self.k = k            for i in range(k):                arr.append(tmp)                tmp = tmp.next            while True:                for i in range(k//2):                    arr[i].val, arr[-1-i].val = arr[-1-i].val, arr[i].val                arr = list(map(self.move, arr))        finally:            return head    def move(self, x):        ret = x        for i in range(self.k):            ret = ret.next        return ret

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