leetcode61

class Solution:
    def constructLink(self,lists):
        n = len(lists)
        if n == 0:
            return None
        if n == 1:
            return ListNode(lists[0])
        
        head = ListNode(lists[-1])
        for i in range(n-2,-1,-1):
            cur = ListNode(lists[i])
            cur.next = head 
            head = cur
        return head
        
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        lists = []
        count = 0
        while head != None:
            lists.append(head.val)
            count += 1
            head = head.next
        if count == 0:
            return None
        k = k % count
        newlist = lists[count-k:] + lists[:count-k]
        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。