LeetCode 142. Linked List Cycle II 判断环入口的位置 C++/Java

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

To represent a cycle in the given linked list, we use an integer poswhich represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

Example 1:

Input: head = [,,,-], pos =
Output: tail connects to node index
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [,], pos =
Output: tail connects to node index
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [], pos = -
Output: no cycle
Explanation: There is no cycle in the linked list.

这题解题的思路在于：在第一次相遇点位置pos，从该位置到环入口Join的距离=从头结点Head到环入口Join的距离

假设环的长度是r，在第一次相遇时，慢指针走过的路程：s=lenA+x，快指针走过的路程：2s=lenA+nr+x，所以：lenA+x=nr，即：LenA=nr-x。

所以在第一次相遇之后，一个指针从head走到join的路程，另一个指针从pos走到join。

方法一（C++）

 ListNode *detectCycle(ListNode *head) {
         ListNode* slow=head,*fast=head;
         while(fast&&fast->next){
             slow=slow->next;
             fast=fast->next->next;
             if(slow==fast)
                 break;
         }
         if(!fast||!fast->next)
             return NULL;
         slow=head;
         while(slow!=fast){
             slow=slow->next;
             fast=fast->next;
         }
         return slow;
     }

（java）：

 ListNode slow=head,fast=head;
         while(fast!=null&&fast.next!=null){
             slow=slow.next;
             fast=fast.next.next;
             if(slow==fast)
                 break;
         }
         if(fast==null||fast.next==null)
             return null;
         slow=head;
         while(slow!=fast){
             slow=slow.next;
             fast=fast.next;
         }
         return slow;
     }