2 Add Two Numbers

Link: https://oj.leetcode.com/problems/add-two-numbers/

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

兩個以鏈表表示的數字，求其的相加值同時以鏈表形式輸出，輸入樣例中：342+465 -> 807。

算法適用於大數的加法，高精度算法等，求解中注意進位即可。

/**
 * 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 val = 0, carry = 0, rval = 0;

		ListNode * re = NULL, *fir=NULL;  // re記錄返回節點，fir記錄插入節點

		while(l2!=NULL && l1!=NULL) {
				
			val = l1->val + l2->val + carry;
			rval = val % 10;
				
			ListNode * cur = new ListNode(rval);
				
			// init
			if(re==NULL && fir==NULL) re = fir = cur;

			else {
				fir->next = cur;
				fir = cur;
			}

			carry = val / 10;

			l1 = l1->next,l2 = l2->next;
		}
		while(carry !=0 || l1!=NULL || l2!=NULL) {

			if(l1 != NULL) {
				val = l1->val + carry;
				l1 = l1->next;
			}
			else if(l2 != NULL){
				val = l2->val + carry;
				l2 = l2->next;
			}
			else val = carry;
			rval = val % 10;

			ListNode * cur = new ListNode(rval);
			fir->next = cur;
			fir = cur;
			carry = val / 10;
		}
		return re;
    }
};