环形链表判断

方法一：

哈希表

时间复杂度：O(n)

空间复杂度：O(n)

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        dict = {}
        while head:
            if head in dict:
                return True
            else:
                dict.setdefault(head,1)
                head = head.next
        return False

方法二：

快慢指针

时间复杂度：O(n)

空间复杂度：O(1)

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        slow = head
        fast = head
        while  fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False