Linked List Cycle II

1. Title

2. Http address

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

3. The question

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?

4. My code(AC)

 1  // Accepted
 2        public static ListNode detectCycle(ListNode head) {
 3            
 4            if( head == null)
 5                return null;
 6            ListNode last ,pre;
 7            last = head;
 8            pre = head.next;
 9            
10            if( pre == last)
11                return head;
12            while( pre != null)
13            {
14                if( pre == last)
15                {
16                    break;
17                }
18                last = last.next;
19                pre = pre.next;
20                if( pre != null)
21                {
22                    pre = pre.next;
23                }else{
24                    pre = null;
25                }
26            }
27            
28            if ( pre == null)
29            {
30                return null;
31            }
32            ListNode second = pre.next;
33            ListNode first = head;
34            pre.next = null;
35            
36            int lenA,lenB;
37           
38            ListNode p = first;
39            lenA = 0 ;
40            while( p != null)
41            {
42                lenA++;
43                p = p.next;
44            }
45            lenB = 0 ;
46            p = second;
47            while( p != null)
48            {
49                lenB++;
50                p = p.next;
51            }
52            
53            if( lenA >= lenB)
54            {
55                int dis = lenA - lenB;
56                while( dis > 0 && first != null)
57                {
58                    first = first.next;
59                    dis--;
60                }
61                
62            }else{
63                int dis = lenB - lenA;
64                while( dis > 0 && second != null)
65                {
66                    second = second.next;
67                    dis--;
68                }
69            }
70            
71            while( first != null && second != null && first != second)
72            {
73                first = first.next;
74                second = second.next;
75            }
76            return second;
77            
78          
79         }

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原文地址：https://www.cnblogs.com/ordili/p/4970027.html