【leetcode】旋转链表

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算法思路

1. 首先确定链表的长度，然后使用k对这个长度取余数，这样避免循环计算
2. 找到倒数第k个节点作为头结点
3. 一直往后遍历到None，连接到首节点
4. 将最后的节点置为None

其实有个更方便的办法：先将链表连接为循环链表（遍历链表长度的时候，就可以连接成循环链表）然后想办法从中间截断，不过时间复杂度是一样的

class Solution:
    def listLength(self, head):
        length = 0
        p = head
        while p:
            length += 1
            p = p.next
        return length
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        if head == None:
            return None
        k = k % (self.listLength(head))
        if k == 0:
            return head
        p = head
        for i in range(k):
            p = p.next
        newHead = head
        while p:
            p = p.next
            newHead = newHead.next
        p = newHead
        while p.next:
            p = p.next
        p.next = head
        p = head
        while p.next != newHead:
            p = p.next
        p.next = None
        return newHead