Given a linked list, return the node where the cycle begins.返回环开始的那个节点 If there is no cycle, return null.没环的话，就返回null
To represent表示 a cycle in the given linked list, we use an integer整数 pos which represents the position位置 (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Note: Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: tail connects to node index 1 Explanation: There is a cycle in the linked list, where tail connects to the second node.Example 2:
Input: head = [1,2], pos = 0 Output: tail connects to node index 0 Explanation: There is a cycle in the linked list, where tail connects to the first node.Example 3:
Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list.Follow up: Can you solve it without using extra space?
把见过的节点丢集合里，下次再遇见就是环的开始(疑问，这里不能链表有重复元素吗？？？)
# 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 """ s = {None} while head not in s: s.add(head) head = head.next return head使用快慢指针的方法，一个纯数学的解法
设环的起始节点为 E，快慢指针从 head 出发，快指针速度为 2，设相交节点为 X，head 到 E 的距离为 H，E 到 X 的距离为 D，环的长度为 L，那么有：快指针走过的距离等于慢指针走过的距离加快指针多走的距离（多走了 n 圈的 L） 2(H + D) = H + D + nL，因此可以推出 H = nL - D，这意味着如果我们让俩个慢指针一个从 head 出发，一个从 X 出发的话，他们一定会在节点 E 相遇
_____ / \ head___________E \ \ / X_____/ class Solution(object): def detectCycle(self, head): slow = fast = head while fast and fast.next: fast = fast.next.next slow = slow.next if slow == fast: break else: return None while head is not slow: head = head.next slow = slow.next return head