[leetcode]Reverse Nodes in k-Group @ Python

原題地址：https://oj.leetcode.com/problems/reverse-nodes-in-k-group/

題意：

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

解題思路：稍微難一點的鏈表反轉題目。需要將鏈表反轉的函數單獨寫成一個函數，這樣看起來會清晰一些。

代碼：

# 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 reverse(self, start, end):
        newhead=ListNode(0); newhead.next=start
        while newhead.next!=end:
            tmp=start.next
            start.next=tmp.next
            tmp.next=newhead.next
            newhead.next=tmp
        return [end, start]
    def reverseKGroup(self, head, k):
        if head==None: return None
        nhead=ListNode(0); nhead.next=head; start=nhead
        while start.next:
            end=start
            for i in range(k-1):
                end=end.next
                if end.next==None: return nhead.next
            res=self.reverse(start.next, end.next)
            start.next=res[0]
            start=res[1]
        return nhead.next