2. 两数相加

符号溢出

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
   long  long   number(ListNode *l){
        long  long  t=1;
         long  long   sum=0;
        while(l!=NULL){
            sum=sum+t*l->val;
            l=l->next;
            t*=10;   
        }
        return sum;
    }
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        // ListNode l;

        cout<<"l1:"<<number(l1);
        cout<<endl;
        cout<<"l2:"<<number(l2);

       
        long long   lNum=number(l1)+number(l2);
     

        ListNode *head=new ListNode (0);
        ListNode *m;
        m=head;
        if(lNum==0) return head ;
 
        while(lNum){
           m->next=new ListNode(lNum%10);
            lNum/=10;
            m=m->next;

        }

    return head->next;

    }
};

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        ret=cur=ListNode(0)
        add=0
        while(l1!=None or l2!=None or add):
            val=(l1.val if l1 else 0)+ (l2.val if l2 else 0)+add
            add=(val)//10
            cur.next=ListNode(val%10)
            cur=cur.next
            l1=l1.next if l1 else None
            l2=l2.next if l2 else None   
        return ret.next                    

c++ 版本

/**
 * 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) {
       ListNode *head=new ListNode(0);
       ListNode *p;
       p=head;
       int add=0;
       while(l1 ||l2 ||add){
           int val= (l1==NULL?0:l1->val)+(l2==NULL?0:l2->val)+add;
           add =val/10;
           p->next=new ListNode(val%10);
           p=p->next;
           l1=(l1==NULL?NULL:l1->next);
           l2=(l2==NULL? NULL :l2->next);


       }
    return head->next;

    }
};

参考