[LeetCode]题解（python）：061-Rotate list

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题目来源

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https://leetcode.com/problems/rotate-list/

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

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题意分析

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Input:a list of node

Output:a list of node shift to right

Conditions:链表右移

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题目思路

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总体思路是链表右移，关键是右移k可能大于实际长度，所以先遍历一次求长度，然后再求模得到真正右移数量

PS：对于链表操作，每次在前面添加一个节点这个方法很好用，特别注重边界条件

PS2：尽量用xrange代替range

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AC代码（Python）

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 __author__ = 'YE'

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

 class Solution(object):
     def rotateRight(self, head, k):
         """
         :type head: ListNode
         :type k: int
         :rtype: ListNode
         """
         if k == 0 or head == None:
             return head

         addFirst = ListNode(0)
         addFirst.next = head
         # move variable
         p = addFirst
         #the length of list
         count = 0
         while p.next != None:
             p = p.next
             count += 1
         p.next = addFirst.next
         #the real step to shift right
         step = count - (k % count)
         p = addFirst.next
         for i in xrange(1, step):
             p = p.next
         head = p.next
         p.next = None

         return head