142. Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

Follow up:
Can you solve it without using extra space?

# 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
        """
        while head:
            if head.val == ‘ ‘:
                return head
            head.val = ‘ ‘
            head = head.next
        return None

我还是修改了链表..

142. Linked List Cycle II

原文：https://www.cnblogs.com/bernieloveslife/p/9745054.html