Leetcode-Easy 141. Linked List Cycle

141. Linked List Cycle

• 描述：
  
  判断一个单链表中是否存在环
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• 思路：
  设置两个指针(fast, slow)，初始值都指向头，slow每次前进一步，fast每次前进二步，如果链表存在环，则fast必定先进入环，而slow后进入环，两个指针必定相遇。(当然，fast先行头到尾部为NULL，则为无环链表)
• 代码

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

class Solution:  
 
    def hasCycle(self, head):  
        # write your code here  
        if head is None:  
            return False  
        if head.next is None:  
            return False  
        slow = head.next  
        fast = head.next  
        if fast.next is not None:  
            fast = fast.next  
        while (slow.next is not None) and (fast.next is not None):  
            if slow == fast:  
                return True  
            slow = slow.next  
            fast = fast.next  
            if fast.next != None:  
                fast = fast.next  
        return False