这一题需要注意一下数据的范围

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/**
 * 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 num1[250],num2[250],len1=0,len2=0;
        for(int i=0;i<250;i++) num1[i]=num2[i]=0;
        while(l1!=NULL)
        {
            num1[++len1]=l1->val;
            l1=l1->next;
        }
        
        while(l2!=NULL)
        {
            num2[++len2]=l2->val;
            l2=l2->next;
        }
        //小技巧，两个数相加，最多只能进一位，因此先让结果多一位
        int len = len1+1;
        if(len1<len2) len = len2+1;
        
        for(int i=1;i<=len;i++)
        {
            num1[i]+=num2[i];
            if(num1[i]/10)
            {
                num1[i+1]++;
                num1[i]=num1[i]%10;
            }
        }
        int cnt = 1;
        while(num1[len]==0 && len>1) len--;
        ListNode* result = new ListNode(num1[cnt++]);
        ListNode* pre = result;
        while(cnt<=len)
        {
            ListNode* node = new ListNode(num1[cnt++]);
            pre->next = node;
            pre = node;
        }
        return result;
    }
};

看了一下题解：
重新写了一下：

/**
 * 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) {
        ListNode* FirHead = new ListNode(0);
        ListNode* p1 = l1,*p2 = l2,*cur = FirHead;
        int carry = 0;
        while(p1 != NULL || p2 != NULL)
        {
            int x = (p1!=NULL)?p1->val:0;
            int y = (p2!=NULL)?p2->val:0;
            int sum = carry + x + y;
            carry = sum /10;
            cur->next = new ListNode(sum%10);
            cur = cur->next;
            if(p1!=NULL) p1=p1->next;
            if(p2!=NULL) p2=p2->next;
        }
        if(carry>0) cur->next = new ListNode(carry);
        return FirHead->next;
    }
};