142. 环形链表 II

解法一：（set）T:O(n)   S:O(n)

用一个set遍历链表，无论成环还是不成环，都会遍历所有的节点一次，区别在于若成环会返回一个访问过的节点，不成环会返回null

# 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
        """
        visited = set()
        node = head
        while node is not None:
            if node in visited:
                return node
            else:
                visited.add(node)
                node = node.next
        return None
        

解法二：Floyd算法

第一阶段：快慢指针（快指针每次移动两个节点，慢指针每次移动一个），记录相遇的节点

第二阶段：找到环的入口，ptr1指向链表的头，ptr2指向相遇的节点

证明过程如下：

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

class Solution(object):
    def getInter(self, head):
        tortoise = head
        hare = head
        while hare is not None and hare.next is not None:
            tortoise = tortoise.next
            hare = hare.next.next
            if tortoise == hare:
                return tortoise
        return None
        
        
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None:
            return head
        interaction = self.getInter(head)
        if interaction is None:
            return None
        ptr1 = head
        ptr2 = interaction
        while ptr1 != ptr2:
            ptr1 = ptr1.next
            ptr2 = ptr2.next
        return ptr1