【Leetcode】2. Add Two Numbers 两数相加

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
代码：

/**
 * 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* cc=new ListNode(-1), *cur=cc;
        int carry=0;
        while(l1||l2){
            int val1 = l1 ? l1->val:0;
            int val2 = l2 ? l2->val:0;
            int sum=val1+val2+carry;
            carry=sum/10;
            cur->next=new ListNode(sum%10);
            cur=cur->next;
            if(l1) l1=l1->next;
            if(l2) l2=l2->next;
        }
        if(carry) cur->next=new ListNode(1);
        return cc->next;
    }
};

参考：
https://www.cnblogs.com/grandyang/p/4129891.html