[LeetCode]25 k个一组翻转链表

题目描述

给出一个链表，每 k 个节点一组进行翻转，并返回翻转后的链表。

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，那么将最后剩余节点保持原有顺序。

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

代码

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

class Solution:
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if k <= 0 or not head or not head.next:
            return head
        dummy = ListNode(-1)
        dummy.next = head
        cur = head
        pre = dummy
        while cur:
            left,right = cur,cur
            for i in range(k-1):
                cur = cur.next
                if cur is None:
                    return dummy.next
                right = cur
            self.nx(pre,left,right)
            cur = left
            pre = left
            cur = cur.next
        return dummy.next
    def nx(self,pre,l,r):
        if not pre or not l or not r or l == r:
            return l
        p ,cur,next = pre,l,l.next
        tmp = r.next
        while cur != tmp:
            cur.next = p
            p = cur
            cur = next
            if next:
                next = next.next
        pre.next = p
        l.next = tmp