如何使用Python实现链表的旋转k个位置操作

# 定义链表节点
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

# 定义链表旋转k个位置的函数
def rotateRight(head: ListNode, k: int) -> ListNode:
    if not head or not head.next or k == 0:  # 链表为空或者只有一个节点或者k=0
        return head

    # 计算链表长度和将k取模
    n, cur = 1, head
    while cur.next:
        n += 1
        cur = cur.next
    k %= n

    if k == 0:  # k%n=0，链表无需旋转
        return head

    # 将链表拆分成两个部分并进行旋转
    cur.next = head  # 连接成环形链表
    for _ in range(n-k):
        cur = cur.next
    new_head = cur.next  # 新的头节点
    cur.next = None  # 新的尾节点

    return new_head  # 返回新的头节点

# 测试
# 输入: 1->2->3->4->5->None，k = 2
# 输出: 4->5->1->2->3->None
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
head.next.next.next.next = ListNode(5)
k = 2
new_head = rotateRight(head, k)
cur = new_head
while cur:
    print(cur.val, end=' ')
    cur = cur.next
# 输出: 4 5 1 2 3