https://leetcode-cn.com/problems/add-two-numbers/

代码如下：

/**
 * 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 len1=0,len2=0;
        vector<int> vec1,vec2;
        while(l1!=NULL){
            vec1.push_back(l1->val);
            len1++;
            l1=l1->next;
        }
        while(l2!=NULL){
            vec2.push_back(l2->val);
            len2++;
            l2=l2->next;
        }
        reverse(vec1.begin(),vec1.end());
        reverse(vec2.begin(),vec2.end());
        int jw=0;
        ListNode *l3=new ListNode(0),*p=new ListNode(0);
        p=l3;
        while(len1>0&&len2>0){
            int key=vec1[len1-1]+vec2[len2-1]+jw;
            if(key>=10){
                key-=10;
                jw=1;
            }else{
                jw=0;
            }
            ListNode *x=new ListNode(key);
            p->next=x;
            p=x;
            p->val=key;
            p->next=NULL;
            len1--,len2--;
        }
        if(len1>0){
            while(len1>0){
                int key=vec1[len1-1]+jw;
                if(key>=10){
                    key-=10;
                    jw=1;
                }else{
                    jw=0;
                }
                ListNode *x=new ListNode(key);
                p->next=x;
                p=x;
                p->val=key;
                p->next=NULL;
                len1--;
            }
        }
        if(len2>0){
            while(len2>0){
                int key=vec2[len2-1]+jw;
                if(key>=10){
                    key-=10;
                    jw=1;
                }else{
                    jw=0;
                }
                ListNode *x=new ListNode(key);
                p->next=x;
                p=x;
                p->val=key;
                p->next=NULL;
                len2--;
            }
        }
        if(jw==1){
            ListNode *x=new ListNode(jw);
            p->next=x;
            p=x;
            p->val=1;
            p->next=NULL;
        }
        return l3->next;
    }
};

原文地址：https://www.cnblogs.com/zzh1582188532/p/12158939.html