一个被干蒙蔽的问题反思

下面这些都是垃圾代码,我的大脑一开始拒绝直接操作节点,因为这些操作都会影响原来链表的数据,我要再想想

#include 
using namespace std;
class ListNode
{
public:
    int val;
    ListNode *nextNodePointer;
    ListNode() : val(0), nextNodePointer(nullptr) {}
    ListNode(int x) : val(x), nextNodePointer(nullptr) {}
    ListNode(int x, ListNode *nextNodePointer) : val(x), nextNodePointer(nextNodePointer) {}
};


void show(ListNode* List)
{
    ListNode* accessNode = List; //访问List用的指针

    while(accessNode != nullptr)
    {
        cout<< accessNode->val << " -> ";
        accessNode = accessNode->nextNodePointer;
    }
    cout<< "nullptr" << endl;
}


ListNode *merge_and_sort_two_list(ListNode* l1 = nullptr, ListNode *l2 = nullptr)
{
        
    
    //方案一:把l2级联到l1上,整体换值排序l1,返回l1
    if(l1 == nullptr)
    {
        l1 = l2;
    }
    else
    {
        ListNode* lastNodePointer = l1;
        while(lastNodePointer->nextNodePointer != nullptr)
            lastNodePointer = lastNodePointer->nextNodePointer;
        lastNodePointer->nextNodePointer = l2;
    }
    //此处需要写一个排序l1的程序
    //返回l1

    
    
    
}

int main(int argc, char* argv[])
{    
    int a[4] = {4,7,2,1};
    int b[4] = {7,5,6,8};
    ListNode *List1 = nullptr;
    ListNode *List2 = nullptr;
    int i = 0;
    while(i < 4)
    {
        List1 = new ListNode(a[i], List1);
        List2 = new ListNode(b[i], List2);
        i = i+1;
    }
    show(List1);
    show(List2);

    ListNode* List = merge_and_sort_two_list(List1,List2);
    show(List);

    return 0;
}

这样的代码有啥用?

ListNode *merge_two_sorted_list(ListNode* l1 = nullptr, ListNode *l2 = nullptr)
{

    if(l1 == nullptr) return l2;
    if(l2 == nullptr) return l1;
    if(l1->val < l2->val)
    {
        l1->nextNodePointer = merge_two_sorted_list(l1->nextNodePointer,l2);
        return l1;
    }
    else
    {
        l2->nextNodePointer = merge_two_sorted_list(l1,l2->nextNodePointer);
        return l2;
    }
}