leetcode#142. Linked List Cycle II

Description

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?
这题回忆满满呀。当初进实验室前，道佳面试我的时候就问了这题。当时临时想的就是最蠢的标记法，或者开数组存储遍历过的节点之类的…显然在不能修改链表和不知道链表有多长的时候，这种做法是不实际的。当时道佳就跟我说了快慢指针的做法，后来也和小嘎嘎讨论过这题，有次还仔细的思考了为啥俩指针速度要差一倍，可惜现在都忘了当初的思考结果了。

Code

# 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
        """
        slow = fast = head
        while slow != None and fast != None:
            slow = slow.next
            fast = fast.next
            if fast == None:
                return None
            else:
                fast = fast.next
            if fast == slow:
                break
        if fast == None:
            return None
        while head != fast:
            head = head.next
            fast = fast.next
        return head

Conclusion

网上的解释很多，随便一搜就有，此处只记录个算法。

http://blog.sina.com.cn/s/blog_6a0e04380101a9o2.html