【题目】

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

【分析】

检查链表内有没有闭环。其他人的思路，两个指针，一个每次跳一个 Node，另一个每次2个 Node，如果存在闭环，最终快指针会追上慢指针，两者相等。如果无闭环，快指针会先到 None。

【代码】

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a ListNode
    # @return a boolean
    def hasCycle(self, head):
        slow = fast = head
        if head == None or head.next == None:
            return False
        else :
            while fast and fast.next :
                slow = slow.next
                fast = fast.next.next
                if fast == slow:
                    return True
            return False