下列python代码用Floyd判断一个单链表中是否存在环,链表的头节点为 head ,即用两个指针在链表上前进:slow 每次走1步,fast 每次走2步,若存在环,fast 终会追上 slow(相遇);若无环,fast 会先到达 nullptr。横线上应填写( )。
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
def hasCycle(head: ListNode) -> bool:
if not head or not head.next:
return False
slow = head
fast = head.next
while fast and fast.next:
if slow == fast:
return True
______
return False
slow = slow.next; fast = fast.next.next
slow.next = slow; fast = fast.next.next
slow = slow.next; fast.next = fast.next.next
slow = slow.next; fast = fast.next