142.md

142. Linked List Cycle II

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/linked-list-cycle-ii/description/

内容描述

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?

解题方案

思路 1 - 时间复杂度: O(N) - 空间复杂度: O(1)

直接快慢指针，跟141一样，。这里注意一下while-else clause的用法就行

beats 99.81%

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                break
        else:
            return None
        while head != slow:
            slow = slow.next
            head = head.next
        return head

如果用可以用字典存储，也可以这样

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        nodemap = {}
        pos = 0
        while head:
            if head in nodemap:
                return head
            nodemap[head] = pos
            pos += 1
            head = head.next
        return None