leedcode 环形链表II

设置快慢指针，确定是否有环。如果有环，确定环中节点的数量。然后再根据快慢指针去找，找到环的入口点。

# 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 head is None or head.next is None:
            return None
        slow = head
        fast = head
        pTemp = None
        while slow and fast:
            slow = slow.next
            fast = fast.next
            if fast:
                fast = fast.next
            else:
                return None
            if slow == fast:
                pTemp = slow
                break
        if pTemp is None:
            return None
        pTemp1 = pTemp.next
        n = 1
        while pTemp1 != pTemp:
            pTemp1 = pTemp1.next
            n = n + 1
        slow = head
        fast = head
        while n > 0:
            fast = fast.next
            n = n - 1
        while slow != fast:
            slow = slow.next
            fast = fast.next
        return slow