1 class Solution: 2 def constructLink(self,lists): 3 n = len(lists) 4 if n == 0: 5 return None 6 if n == 1: 7 return ListNode(lists[0]) 8 9 head = ListNode(lists[-1]) 10 for i in range(n-2,-1,-1): 11 cur = ListNode(lists[i]) 12 cur.next = head 13 head = cur 14 return head 15 16 def rotateRight(self, head: ListNode, k: int) -> ListNode: 17 lists = [] 18 count = 0 19 while head != None: 20 lists.append(head.val) 21 count += 1 22 head = head.next 23 if count == 0: 24 return None 25 k = k % count 26 newlist = lists[count-k:] + lists[:count-k] 27 return self.constructLink(newlist)
先将原来的链表展开，存储到数组中，并计算数组长度。
再将数组按照k值翻转，重新排好顺序。
最后使用尾插法，重建链表。
将复杂问题分开步骤处理，可以使代码更清晰。