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;
}
};