Leetcode 2. Add Two Numbers（鏈表求和）

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：

兩個數字求和，數字用鏈表表示，每一個結點代表一位。鏈表順序與數字順序相反，即表頭存放數字的最低位。

/**
 * 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;
    }
};