61. 旋转链表

61. 旋转链表
->每个节点都要移动，因此想到连接到一起
    1.找尾节点，形成环形链表
    2.尾节点移动length-k步(又移k步==左移length-k步)
    3.找到头节点，断开头尾连接
-<代码：
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
demo1:
class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        if not head or not head.next:
            return head # 特例判断
        tail = head # tail指向head
        while tail.next:
            length+=1
            tail = tail.next #记录链表长度
        tail.next=head #头尾相连
        k %=length # 确定移动后的位置移动
        for _ in range(length-k):
            tail = tail.next
        head = tail.next
        tail.next = None #断开链表
        return head
demo2:
class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        # 若链表为空，返回head
        if not head:
            return head
        # 获取链表的长度
        n = 0
        cur = head
        while cur:
            n += 1
            cur = cur.next
        # 若k的长度大于n，将k对n求余
        k = k % n
        # 若k==0表示循环了一整遍，直接返回head
        if k == 0:
            return head
        # 创建快慢指针
        slow = fast = head
        # fast指针先走k步
        while k:
            fast = fast.next
            k -= 1
        # 让fast指针走到队尾
        while fast.next:
            fast = fast.next
            slow = slow.next
        # 此时show.next为新的链表头
        new_head = slow.next
        # 断开slow.next
        slow.next = None
        # 链表首位相接
        fast.next = head
        # 返回新的链表头
        return new_head