2两数相加

#include
using namespace std;
struct ListNode {
    int val;
    ListNode* next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode* next) : val(x), next(next) {}
};   //定义链表节点
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int carry = 0;//初始化进位值
        ListNode* head = nullptr;
        ListNode* tail = nullptr; //初始化，链表头尾都是空
        while (l1 || l2)
        {
            int n1 = l1 ? l1->val : 0;
            int n2 = l2 ? l2->val : 0;//读取当前位置两个数组的值
            int sum = n1 + n2 + carry;
            if (!head) //如果表头为空，即整个链表为空
            {
                head = tail = new ListNode(sum % 10);//新建一个节点
            }
            else//链表不为空
            {
                tail->next = new ListNode(sum % 10);
                tail = tail->next;//更新链表尾部
            }
            carry = sum / 10;//新的进位值
            if (l1)
            {
                l1 = l1->next;
            }
            if (l2)
            {
                l2 = l2->next;
            }//如果l1,l2没有更新完就继续
        }
        if (carry > 0)//l1,l2更新完后如果有进位就再新建一个节点
        {
            tail->next = new ListNode(carry);
        }
        return head;//head为数组基地址，这里即为返回数组
    }
};
void print(ListNode* head)//输出结果链表
{
    ListNode* p;
    p = head;
    while (p != NULL)
    {
        cout << p->val << " ";
        p = p->next;
    }
}
int main()
{
    ListNode* l1 = new ListNode(2);
    ListNode* l2 = new ListNode(5);
    l1->next = new ListNode(4);
    l2->next = new ListNode(6);
    l1->next->next = new ListNode(3);
    l2->next->next = new ListNode(4);//构建l1,l2
    Solution test;
    ListNode *a = test.addTwoNumbers(l1, l2);
    print(a);
    return 0;
}