Python中链表的倒数第k个节点查找方法

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

def find_kth_to_tail(head, k):
    if head is None or k <= 0:
        return None
    p1 = head
    p2 = head
    for i in range(k-1):
        if p2.next is not None:
            p2 = p2.next
        else:
            return None
    while p2.next is not None:
        p2 = p2.next
        p1 = p1.next
    return p1

# 测试示例
if __name__=='__main__':
    head = ListNode(0)
    node1 = ListNode(1)
    node2 = ListNode(2)
    node3 = ListNode(3)
    node4 = ListNode(4)
    head.next = node1
    node1.next = node2
    node2.next = node3
    node3.next = node4
    # 链表为 0 -> 1 -> 2 -> 3 -> 4
    print(find_kth_to_tail(head, 2).val)   # 输出 3