Leetcode 2. Add Two Numbers（链表求和）

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

思路：

两个数字求和，数字用链表表示，每一个结点代表一位。链表顺序与数字顺序相反，即表头存放数字的最低位。

/** * 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 carry=0;        ListNode* l3 = new ListNode(0);        ListNode* head = l3;        while(l1||l2) {            int tmp = carry;            if(l1) tmp+=l1->val,l1=l1->next;            if(l2) tmp+=l2->val,l2=l2->next;            head->next = new ListNode(tmp%10);            carry=tmp/10;            head=head->next;        }        if(carry) head->next = new ListNode(carry);        return l3->next;    }};