LeetCode----Linked List Cycle

Linked List Cycle

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

Follow up:

Can you solve it without using extra space?

分析：

判断一个链表是否含有环。

使用弗洛伊德循环检测法。时间复杂度为O（n）。

算法详解如下：

Floyd's Cycle Detection Algorithm (The Tortoise and the Hare)

How do you determine if your singly-linked list has a cycle? In the late 1960s, Robert W. Floyd invented an algorithm that worked in linear (O(N)) time. It is also called Floyd's cycle detection algorithm.

The easiest solution to the cycle detection problem is to run through the list, keeping track
of which nodes you visit, and on each node check to see if it is the same as any of the previous nodes. It's pretty obvious that this runs in quadratic (O(N^2)) time... not very efficient, and actually more complicated than this one.

The Tortoise and the Hare is possibly the most famous cycle detection algorithm, and is surprisingly straightforward. The Tortoise and the Hare are both pointers, and both start at the top of the list. For each iteration, the Tortoise takes one step and the
Hare takes two. If there is a loop, the hare will go around that loop (possibly more than once) and eventually meet up with the turtle when the turtle gets into the loop. If there is no loop, the hare will get to the end of the list without meeting up with
the turtle.

Why can't you just let the hare go by itself? If there was a loop, it would just go forever; the turtle ensues you will only take n steps at most.

For a somewhat more efficient algorithm, check out Brent's Cycle Detection Algorithm (The Teleporting Turtle).



伪码：

<pre name="code" class="python"> 

tortoise = top
hare = top

forever:
if hare == end :
return 'No Loop Found'
hare = hare.next

if hare == end :
return 'No Loop Found'
hare = hare.next

tortoise = tortoise.next

if hare == tortoise:
return 'Loop Found'

我的代码：

class Solution(object):
def hasCycle(self, head):
"""
:type head: ListNode
:rtype: bool
"""
q = p = head
while q and q.next:
p = p.next
q = q.next.next
if p == q:
return True
return False