forked from CodingVault/LeetCodeInPython
-
Notifications
You must be signed in to change notification settings - Fork 0
/
linked_list_cycle_ii.py
71 lines (59 loc) · 1.68 KB
/
linked_list_cycle_ii.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
#!/usr/bin/env python
# encoding: utf-8
"""
linked_list_cycle_ii.py
Created by Shengwei on 2014-06-30.
"""
# https://oj.leetcode.com/problems/linked-list-cycle-ii/
# tags: medium, linked-list, logic
"""
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up:
Can you solve it without using extra space?
"""
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
############ V2 ############
class Solution:
# @param head, a ListNode
# @return a list node
def detectCycle(self, head):
if head is None or head.next is None:
return
walker, runner = head, head
while runner and runner.next:
walker = walker.next
runner = runner.next.next
if runner == walker:
break
if runner != walker:
return
walker = head
while runner != walker:
runner = runner.next
walker = walker.next
return walker
############ V1 ############
class Solution:
# @param head, a ListNode
# @return a list node
def detectCycle(self, head):
if head is None:
return None
walker = head.next
if walker is None:
return None
runner = head.next.next
while runner != walker:
if runner is None or runner.next is None:
return None
runner = runner.next.next
walker = walker.next
walker = head
while runner != walker:
runner = runner.next
walker = walker.next
return walker