141/142. Linked List Cycle I/II

141. 判断链表是否有环.
     用快慢指针，注意点在于考虑清楚链表长度为0，1，2，3的情况。while循环判断条件建议用fast != slow

    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if not head or not head.next:
            return False

        slow = head.next
        fast = head.next.next

        while fast != slow:
            if not fast or not fast.next:
                return False
            slow = slow.next
            fast = fast.next.next
        return True

142. 找到入环点。注意点在于初始化的时候让slow先走一步，fast先走两步，始终保持1:2的关系。

    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return None

        slow = head.next
        fast = head.next.next

        while fast != slow:
            if not fast or not fast.next:
                return None
            fast = fast.next.next
            slow = slow.next

        cur = head
        while cur != slow:
            cur = cur.next
            slow = slow.next
        return cur