leetcode 142. 环形链表 II (python3)

142. 环形链表 II
给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 null。

相遇之后，fast 和slow都用一倍速跑，相遇的地方便是起点

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

class Solution:
    def detectCycle(self, head: ListNode) -> ListNode:
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                fast = head#将fast放到起始位置
                break
        if fast is None or fast.next is None:
            return None
        while fast != slow:
            slow = slow.next
            fast = fast.next
        return slow