判断两个链表是否合并

public static <T> T getMerge(Node<T> link1, Node<T> link2){
        Node<T> current1 = link1;//current1先遍历链表1再遍历链表2
        Node<T> current2 = link2;//current2先遍历链表2再遍历链表1
        // 找到共同节点，或分别遍历完成，结束
        while(!current1.value.equals(current2.value)&&(current1.next!=null && current2.next!=null)){
            if(current1.next == null){
                current1 = link2;
            }else{
                current1 = current1.next;
            }
            if(current2.next == null){
                current2 = link1;
            }else{
                current2 = current2.next;
            }
        }
        if(current1.value.equals(current2.value)){// 找到共同节点
            return current1.value;
        }else{// 遍历结束也没有找到共同节点
            return null;
        }
    }
    public static class Node<T>{
        T value;
        Node<T> next;
        Node(T value){
            this.value = value;
        }
    }

@Test
    public void getMerge() {
        /**
         * a->b->c->x->y->z
         * d->e->f->x->y->z
         * 返回x
         */
        MergeOfLinkList.Node<String> link1 = createList(new String[]{"a","b","c","x","y","z"});
        MergeOfLinkList.Node<String> link2 = createList(new String[]{"d","e","f","x","y","z"});
        assertEquals("x", MergeOfLinkList.getMerge(link1, link2));
        /**
         * a->b->c->x->y->z
         * d->e->f->g
         * 返回 null
         */
        MergeOfLinkList.Node<String> link3 = createList(new String[]{"d","e","f","g"});
        assertNull(MergeOfLinkList.getMerge(link1, link3));
    }