解题思路

因为是两个升序链表，用两个指针去遍历各自的链表，直到任意指针为空（某一个链表遍历完毕）；
如果a_head的值大于b_head的值, 则b_head++(去寻找能等于a链表的节点)
如果a_head的值小于b_head的值, 则a_head++(去寻找能等于b链表的节点)
如果a_head的值等于b_head的值, 则a_head++，b_head++打印公共部分

代码(c++)

void sol(list_node * a_head, list_node * b_head)
{
    //////在下面完成代码
    // 如果链表为空
    if ( a_head == NULL || b_head == NULL ){ cout << "no recover area!";}
    list_node* a_pointer = a_head;
    list_node* b_pointer = b_head;
    while( a_pointer!=NULL && b_pointer!=NULL ){
        if( a_pointer->val == b_pointer->val ){
            cout << a_pointer->val << " ";
            a_pointer = a_pointer->next;
            b_pointer = b_pointer->next;
        }
        else if( a_pointer->val >= b_pointer->val ){
            b_pointer = b_pointer->next;
        }
        else{
            a_pointer = a_pointer->next;
        }
    }
}