题目:

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

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

代码:oj在线测试通过 Runtime: 416 ms

 # 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):
if head is None or head.next is None:
return False p1 = ListNode(0)
p1.next = head
p2 = ListNode(0)
p2.next = head result = False
while p1 is not None and p2 is not None:
if p1 == p2:
result = True
break
else:
p1 = p1.next
p2 = p2.next
if p2 is not None:
p2 = p2.next return result

思路:

这是一道经典的题 关键点是快慢指针

p1是慢指针,一次走一步;p2是快指针,一次走两步;如果有循环,则快慢指针一定会在某一时刻遇上。

有个问题比较关键:为啥进入循环后,快指针一定能在某一时刻跟慢指针踩在同一个点儿上呢?

小白觉得可以如下解释:

假设现在快慢指针都在循环当中了,由于循环是个圈,则可以做如下的等价:“慢指针一次走一步,快指针一次走两步” 等价于 “慢指针原地不动,快指针一次走一步”这个其实跟物理学中的相对运动原理差不多。

欢迎高手来拍砖指导。

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