两个单向链表相交问题

单链表可能有环，也可能无环。给定两个单链表的头节点 head1和head2，这两个链表可能相交，也可能 不相交。请实现一个函数， 如果两个链表相交，请返回相交的第一个节点；如果不相交，返回null 即可。 要求：如果链表1 的长度为N，链表2的长度为M，时间复杂度请达到 O(N+M)，额外空间复杂度请达到O(1)。

public class FindForstNode {
    public class Node{
        public int value;
        public Node next;
        public Node(int data){
            this.value = data;
        }
    }
    public static Node getFirstNode(Node head1,Node head2){
        if(head1 == null || head2 == null){
            return null;
        }
        Node loop1 = getLoopNode(head1);//有环的结点，没环为null
        Node loop2 = getLoopNode(head2);
        if(loop1 == null && loop2 == null){
            return noLoop(head1,head2);
        }
        if(loop1 != null && loop2 != null){
            return bothLoop(head1,loop1,head2,loop2);
        }
        return null;
    }
    //快慢指针判断是否有环
    public static Node getLoopNode(Node head){
        if(head == null || head.next == null || head.next.next == null){
            return null;
        }
        Node n1 = head.next;
        Node n2 = head.next.next;
        while(n1 != n2){
            if(n2.next == null || n2.next.next == null){
                return null;
            }
            n1 = n1.next;
            n2 = n2.next.next;
        }
        n2 = head;
        while(n1 != n2){
            n1 = n1.next;
            n2 = n2.next;
        }
        return n1;
    }
    //无环
    public static Node noLoop(Node head1,Node head2){
        if(head1 == null || head2 == null){
            return null;
        }
        Node cur1 = head1;
        Node cur2 = head2;
        int n = 0;
        while(cur1 != null){
            n++;
            cur1 = cur1.next;
        }
        while(cur2 != null){
            n--;
            cur2 = cur2.next;
        }
        //不相交
        if(cur1 != cur2){
            return null;
        }
        cur1 = n > 0 ? head1 : head2;
        cur2 = cur1 == head1 ? head2 : head1;
        n = Math.abs(n);
        while(n != 0){
            cur1 = cur1.next;
            n--;
        }
        while(cur1 != cur2){
            cur1 = cur1.next;
            cur2 = cur2.next;
        }
        return cur1;
    }
    //有环
    public static Node bothLoop(Node head1,Node loop1,Node head2, Node loop2){
        Node cur1 = null;
        Node cur2 = null;
        if(loop1 == loop2){
            cur1 = head1;
            cur2 = head2;
            int n = 0;
            while(cur1 != loop1){
                n++;
                cur1 = cur1.next;
            }
            while(cur2 != loop2){
                n--;
                cur2 = cur2.next;
            }
            cur1 = n > 0 ? head1 : head2;
            cur2 = cur1 == head1 ? head2 : head1;
            n = Math.abs(n);
            while(n != 0){
                cur1 = cur1.next;
                n--;
            }
            while(cur1 != cur2){
                cur1 = cur1.next;
                cur2 = cur2.next;
            }
            return cur1;
        }else{
            cur1 = loop1.next;
            while(cur1 != loop1){
                if(cur1 == loop2){
                    return loop1;
                }
                cur1 = cur1.next;
            }
            return null;
        }
    }
}