leetcode -- Reverse Nodes in k-Group -- 经典题目，要重写

https://leetcode.com/problems/reverse-nodes-in-k-group/

思路很简单，复习的时候再写一遍

自己的code，自己可以写出来，有点慢，多写几次。在reverseList的时候，要注意先把tail.next置为None

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

class Solution(object):

    def reverseList(self, head, tail):
        last, cur = None, head
        new_tail = head
        tail.next = None#这里注意进来先把tail.next置为None
        while cur:
            tmp = cur.next
            cur.next = last
            last = cur
            cur = tmp
        new_head = last
        return new_head, new_tail


    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head: return None
        if k == 1: return head

        cur = head
        dummy = ListNode(0)
        last_tail = dummy
        while cur:

            count = 0
            h = cur
            pre = None
            while cur and count < k:
                pre = cur
                cur = cur.next
                count += 1
            t = pre
            if count == k:
                new_h, new_t = self.reverseList(h, t)
                print (h.val, t.val, new_h.val, new_t.val)
            else:
                new_h, new_t = h, t
            last_tail.next = new_h
            last_tail = new_t

        return dummy.next