输入:1->2->3->4->5->6->7->3(3为链表中的第三个节点)
输出:True
思路:
快慢法来求解,slow走一步,fast走2步,从head出发,如果带环的话,必定相遇。
代码
# 判断链表是否有环
def is_Loop(head):
if head is None or head.next is None:
return False
slow = head
fast = head
while fast is not None and fast.next is not None:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False
# 获得相遇的点
def _get_meet_Node(head):
if head is None or head.next is None:
return None
slow = head
fast = head
while fast is not None and fast.next is not None:
slow = slow.next
fast = fast.next.next
if slow == fast:
return slow
return None
# 获得环入口的节点
def get_Loop_Node(head, meetNode):
first = head
second = meetNode
while first != second:
first = first.next
second = second.next
return first
# 测试
# 构造带环链表
link = LinkedList()
link.add(1)
link.add(2)
link.add(3)
link.add(4)
link.add(5)
link.add(6)
new_node = Node(7)
cur = link.head
entry_node = link.head
while cur.next:
cur = cur.next
while entry_node.data != 3:
entry_node = entry_node.next
cur.next = new_node
new_node.next = entry_node
# 调用函数测试
print("is_Loop(link.head): ", is_Loop(link.head))
meet_Node = _get_meet_Node(link.head)
print("_get_meet_Node(link.head): ", meet_Node.data)
print("get_Loop_Node(link.head, meet_Node).data:", get_Loop_Node(link.head, meet_Node).data)
# 结果
is_Loop(link.head): True
_get_meet_Node(link.head): 6
get_Loop_Node(link.head, meet_Node).data: 3
第三个函数,通过计算,可以知道链表头到环入口的距离=(n-1)* 环长 + 相遇点到环入口的距离,n为相遇前走的圈数。从而表头与相遇点各自设一个指针,每次各走一步,两指针必定相遇,且在环入口点。(参考猿媛之家的python程序员面试算法宝典)