2019-05-18LeetCode141. 环形链表

25min，双指针，对链表的处理，关键点在于a.next 的时候 a 不能是None，特殊情况head 为空

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if(head!=None):pos=head.next
        else:return False  # head=None
        while(pos!=head):
            if(head!=None):
                head=head.next
            else:return False
            if(pos!=None and pos.next!=None):
                pos=pos.next.next
            else:return False
        return True

使用try catch简直优雅，因为所有的异常情况都表示无环！excellent！
Easier to ask for forgiveness than permission."

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        try:
            fast=head.next
            while(fast!=head):
                fast=fast.next.next
                head=head.next
            return True
        except:
            return False