python检查链表是否存在环

思想创建两个游标,一快一慢,相遇后慢游标回到链表头节点,快游标放慢,然后快慢游标会在链表环的头部相遇
class Node(object):
    '''创建一个节点'''
    def __init__(self,item):
        self.item = item
        self.next = None
def findloop(head):
    slow = head
    fast = head
    loopExist = False
    # 创建一个标记
    if head == None:
        return False
    # 如果头节点为空,直接返回
    while fast.next!=None and fast.next.next != None:
        # 只要两个游标不为空就一直循环
        # 两个游标一快一慢,终会相遇
        fast = fast.next.next
        # 快游标
        slow = slow.next
        # 慢游标
        if slow == fast:
            # 游标相遇,返回True,并跳出循环
            loopExist=True
            break
    if loopExist ==  True:
        # 如果相遇,说明有环
        slow = head
        # 将慢游标从头节点重新开始
        while slow != fast:
            # 两个游标不相等就一直循环
            slow = slow.next
            fast = fast.next
        return slow.item
        # 相等以后,返回环的头节点
    return False

node1 = Node(1)
node2 = Node(2)
node3 = Node(3)
node4 = Node(4)

node1.next = node2
node2.next = node3
node3.next = node4

if findloop(node1):
    print('存在环')
    print(findloop(node1))
else:
    print("不存在环结构")

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转载自blog.csdn.net/qlzy_5418/article/details/85240669
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