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 = Noneclass Solution(object):def rotateRight(self, head, k):""":type head: ListNode:type k: int:rtype: ListNode"""node = headcnt = 0# 计算链表长度while node:cnt += 1node = node.next# 保持原地不动：无需移动的情况if cnt <= 1 or k%cnt ==0:return headk %= cntj = cnt - kdummy = ListNode(0)cur = dummynode = headwhile j > 0:cur.next = ListNode(node.val)cur = cur.nextnode = node.nextj -= 1res = nodewhile node.next:node = node.next        node.next = dummy.nextreturn res

大神版本:

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

 

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