061 rotate list
61. Rotate List
题目: https://leetcode.com/problems/rotate-list/
难度:
Medium
- k可能比list的size大，需要做一个取余准备
- 计算list size的同时把tail也记录下来，方便之后把tail的next指向原本的head
- 利用之前的到末端的kth node
AC 代码
class Solution(object):
def rotateRight(self, head, k):
if head == None or k == 0 :
return head
cur = head
size = 1
while cur.next:
size += 1
cur = cur.next
tail = cur
k = k % size
p = self.findKth(head,k)
tail.next = head
head = p.next
p.next = None
return head
def findKth(self,head, k):
dummy = ListNode(-1)
dummy.next = head
p = dummy
q = dummy
for i in range(k):
q = q.next
while q.next:
p = p.next
q = q.next
return p
