翻转链表中相邻的k个节点

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