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 = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
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;
    }