Given the head of a linked list, rotate the list to the right by k places.
给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。
Input: head = 1,2,3,4,5, k = 2
Output: 4,5,1,2,3
class Solution:
def rotateRight(self, head, k):
if not head or not head.next: return head
求链表长度
_len = 0
cur = head
while cur:
_len += 1
cur = cur.next
对长度取模
k %= _len
if k == 0: return head
让 fast 先向后走 k 步
fast, slow = head, head
while k:
fast = fast.next
k -= 1
此时 slow 和 fast 之间的距离是 k；fast 指向第 k+1 个节点
当 fast.next 为空时，fast 指向链表最后一个节点，slow 指向倒数第 k + 1 个节点
while fast.next:
fast = fast.next
slow = slow.next
newHead 是倒数第 k 个节点，即新链表的头
newHead = slow.next
让倒数第 k + 1 个节点 和 倒数第 k 个节点断开
slow.next = None
让最后一个节点指向原始链表的头
fast.next = head
return newHead