标签：没有 node solution 翻转 int ret ini elf lse
示例：
输入：1->2->3->4->5
k=2 输出：2->1->4->3->5
k=3输出：3->2->1->4->5
Python解决方案1：
# 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 """ if k == 1: return head out_head = ListNode(0) out = out_head while head: i = 0 part = ListNode(0) new = part while i < k and head: part.next = head head = head.next part = part.next i += 1 if i == k: part.next = None out.next, last = self.reverse(new.next) out = last else: out.next = new.next return out_head.next def reverse(self,head): prev = None while head: head.next,prev,head = prev,head,head.next last = prev while last.next: last = last.next return prev,last
Python解决方案2：
# 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
"""
if k == 1:
return head
out = ListNode(1)
out_head = out
while head:
prev = None
i = 0
# 翻转k个相邻的节点
while i < k and head:
head.next,prev,head = prev,head,head.next
if i == 0:
last = prev
i += 1
# 将翻转后的短链表连接到要输出的链表后面
if i == k:
out_head.next = prev
out_head = last
# 最后一次翻转如果没有k个节点，则再翻转一次恢复原来的顺序
else:
re_pre = None
while prev:
prev.next,re_pre,prev = re_pre,prev,prev.next
out_head.next = re_pre
return out.next
标签：没有 node solution 翻转 int ret ini elf lse
原文地址：https://www.cnblogs.com/wenqinchao/p/10676959.html
