【Leetcode】21.合并两个有序链表C++

#include "iostream"

using namespace std;

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution
{
public:
    void print(ListNode *head)
    {
        ListNode *p = head;
        while (p != NULL)
        {
            cout << p->val << " ";
            p = p->next;
        }
        cout << endl;
    }
    // l1: 2
    // l2: 1
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2)
    {
        if (l1 == NULL)
        {
            return l2;
        }
        else if (l2 == NULL)
        {
            return l1;
        }

        ListNode *p1, *p2, *s;
        if (l1->val <= l2->val)
        {
            p1 = l1;
            p2 = l2;
        }
        else
        {
            p1 = l2;
            p2 = l1;
            l1 = l2;
        }

        while (p2 != NULL)
        {
            if(p1->next==NULL)
            {
                p1->next = p2;
            }
            if (p1->next->val < p2->val)
            {
                p1 = p1->next;
            }
            else
            {
                s = p2->next;
                p2->next = p1->next;
                p1->next = p2;
                p2 = s;
            }
        }

        return l1;
    }
};

int main(int argc, char const *argv[])
{
    ListNode *l1, *l2, *p;
    int x;
    Solution so;

    cout << "input l1:" << endl;
    cin >> x;
    p = new ListNode(x);
    l1 = p;
    while (true)
    {
        cin >> x;
        p->next = new ListNode(x);
        p = p->next;
        if (cin.get() == '\n')
            break;
    }

    cout << "input l2:" << endl;
    cin >> x;
    p = new ListNode(x);
    l2 = p;
    while (true)
    {
        cin >> x;
        p->next = new ListNode(x);
        p = p->next;
        if (cin.get() == '\n')
            break;
    }

    so.mergeTwoLists(l1, l2);

    so.print(l1);

    return 0;
}