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[leetcode]Reverse Nodes in k-Group @ Python
原题地址：https://oj.leetcode.com/problems/reverse-nodes-in-k-group/
题意：
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
解题思路：稍微难一点的链表反转题目。需要将链表反转的函数单独写成一个函数，这样看起来会清晰一些。
简单插图如下：
# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: # @param head, a ListNode # @param k, an integer # @return a ListNode def reverseKGroup(self, head, k): #迭代版，时间复杂度O(n)，空间复杂度O(1) # Need help if head == None: return head dummy = ListNode(0) dummy.next = head start = dummy # start =0. Given this linked list: 1->2->3->4->5 # For k = 2, you should return: 2->1->4->3->5 while start.next: end = start # end = 0 for i in range(k-1): end = end.next # end = 1 if end.next == None: return dummy.next (start.next, start)=self.reverse(start.next, end.next) # (start.next=3, start=1)=self.reverse(start.next=1, end.next=2) return dummy.next def reverse(self, start, end): dummy = ListNode(0) dummy.next = start while dummy.next != end: tmp = start.next start.next = tmp.next tmp.next = dummy.next dummy.next = tmp #start.next, start.next.next, dummy.next = start.next.next, dummy.next, start.next # The above line is wrong! But WHY? return (end, start)"""Let k = 2, List be: 0 --> 1 --> 2 --> 3 --> 4 --> 5 Initial State: start start.next | | V V 0 --> 1 --> 2 --> 3 --> 4 --> 5 After dealing with first group (i.e., first k nodes) start start.next | | V V 0 --> 2 --> 1 --> 3 --> 4 --> 5"""
[leetcode]Reverse Nodes in k-Group @ Python
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