#2. Add Two Numbers-优快云博客

本文介绍了一个使用链表存储数字并实现两个链表相加的方法。通过遍历链表节点进行逐位相加，并处理进位逻辑，最终返回新的链表结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
       int digit_Num,extra_Num=0,random;
        ListNode *tou=NULL,*p1,*p2,*temp;
        while(l1!=NULL&&l2!=NULL){
            digit_Num = (l1->val+l2->val+extra_Num)%10;
            //cout<<"digit_Num = "<<digit_Num<<endl;
            if(tou==NULL){
                //new malloc还是有相似的地方
                //p2 = p1=tou = new ListNode(digit_Num+extra_Num);
                p2 = p1 = tou =(ListNode*)new ListNode(digit_Num+extra_Num);
                //cout<<"tou "<<tou->val<<endl;
            }else{
                p1 = (ListNode*)new ListNode(digit_Num);
                p2->next = p1;
                p2 = p1;
            }

            extra_Num = (l1->val+l2->val+extra_Num)/10;
            //cout<<"extra_Num = "<<extra_Num<<endl;

            //temp = l1;
            l1 = l1->next;
            //free(temp);
            //temp = l2; 可能是栈上的内存 不能瞎释放
            //free(temp);
            l2 = l2->next;

        }
        //cout<<" Here extra_Num = "<<extra_Num<<endl;

        if(l1==NULL&&l2==NULL){
            //cout<<"3  code here"<<"ex = "<<extra_Num<<endl;
            if(extra_Num!=0)
                p2->next = (ListNode*)new ListNode(extra_Num);
        }else if(l1==NULL){
            temp = l2;
            while(temp!=NULL&&extra_Num!=0)
            {
                random = temp->val;
                //cout<<"temp->val = "<<temp->val<<endl;
                temp->val = (extra_Num + temp->val)%10;
            //  cout<<"temp->val = "<<temp->val<<endl;
                extra_Num = (extra_Num+random)/10;
                temp = temp->next;
            }
            //cout<<"ex = "<<extra_Num<<endl;
            if(extra_Num!=0){
                temp = l2;
                //cout<<"1 Code come to here"<<endl;
                while(temp->next!=NULL)
                    temp = temp->next;
                temp->next = (ListNode*)new ListNode(extra_Num);
            }

            p2->next = l2;
        }else if(l2==NULL){
            temp = l1;
            while(temp!=NULL&&extra_Num!=0)
            {
                random = temp->val;
                temp->val = (extra_Num + temp->val)%10;
                extra_Num = (extra_Num+random)/10;
                temp = temp->next;
            }
        //  cout<<"ex = "<<extra_Num<<endl;
            if(extra_Num!=0){
                temp = l1;
            //  cout<<"2 Code come to here"<<endl;
                while(temp->next!=NULL)
                    temp = temp->next;
                temp->next = (ListNode*)new ListNode(extra_Num);
            }

            p2->next = l1;
        }

        return tou;
    }
};

无限大的数字相加问题，一般会模拟加法运算，同时利用栈来提高效率，本题是一样的，倒序存放的链表和栈是一样的，操作复杂了一点，以上代码可以做一点优化，但是算法本身没有办法提速。