Leetcode #142 Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return

null

.

Note: Do not modify the linked list.

Follow up:

Can you solve it without using extra space?

Difficulty: Medium

如果不用额外的space，这里有一个小技巧。

假设我们在第一次p = pNext时，p走了N步，pNext走了2N步，两个指针的步数之差正好就是一个cycle的步数，也就是2N-N = N。

设进入这个循环需要X步，那么此时指针p和pNext离cycle的起点正好差X步（可以画个图想想）。所以，让p回到head，同时开始走，每次都只走一步，下一次相遇就是cycle的入口。

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

class Solution(object):
def detectCycle(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
flag = 0
p = head
pNext=head

while (p is not None) and (pNext is not None) and flag==0:
if p.next is None or pNext.next is None:
return None
p = p.next
pNext = pNext.next.next
if p==pNext:
flag = 1
if flag == 0:
return None
p = head;
while p!=pNext:
p = p.next
pNext = pNext.next
return p