[leetcode]Reverse Nodes in k-Group

题目：

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

For 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

描述：每K个节点一组倒序

solution by python:

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

class Solution:
    # @param head, a ListNode
    # @param k, an integer
    # @return a ListNode
    def reverseKGroup(self, head, k):
        p = head
        for i in range(k-1):
            if p==None: return head
            p = p.next
        if p == None: return head
        h = self.reverseKGroup(p.next, k)
        self.reverse(head, p)
        head.next = h
        return p
    def reverse(self, start, end):
        tail = None
        while start != end:
            tmp = start.next
            start.next = tail
            tail = start
            start = tmp
        end.next = tail