Leetcode 61. 旋转链表 Python

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我的版本：

# 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
        """

        node = head
        cnt = 0
        # 计算链表长度
        while node:
            cnt += 1
            node = node.next
        # 保持原地不动：无需移动的情况
        if cnt <= 1 or k%cnt ==0:
            return head
        k %= cnt
        j = cnt - k
        dummy = ListNode(0)
        cur = dummy
        node = head
        
        while j > 0:
            cur.next = ListNode(node.val)
            cur = cur.next
            node = node.next
            j -= 1

        res = node
        while node.next:
            node = node.next        
        node.next = dummy.next
        return res

大神版本:

    def rotateRight(self, head, k):
        if k == 0 or head is None:
            return head
        
        dummy=ListNode(0)
        dummy.next=head
        p=dummy
        length=0
        
        while p.next:                       # 得到list的长度
            length+=1
            p=p.next
        
        p.next=dummy.next                   # 指针指向head
        
        step=length-(k%length)              # 得到新head的位置
        
        while step>0:                       # 到新head的位置
            step-=1
            p=p.next
        
        dummy.next=p.next                   # 变化
        p.next=None                         #
        
        return dummy.next