[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