LeetCode142：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?

142和142合并就是完整的剑指Offer_编程题55：链表中环的入口结点。

runtime打败了99.78%的人呀！

# 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
        """
        if not head:
            return None
        fast = slow = head 
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                fast = head
                while fast and slow:
                    if fast == slow:
                        return slow
                    fast = fast.next
                    slow = slow.next
        return None