LeetCode Add Two Numbersの2つのチェーンテーブルの数値加算

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
無限加算に相当し、チェーンテーブルでデータを格納します.
この問題はチェーンテーブルの操作とシミュレーション加算を試験することです.
1つ目の方法:
これは,2つの秩序チェーンテーブルを1つの秩序チェーンテーブルに統合するプログラム構造と同じである.
 

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
		int carry = 0;
		int sum = 0;
		ListNode *sumList = new ListNode(-1);
		ListNode *cur = sumList;
		int l1val, l2val;
		while (l1 && l2)
		{
			l1val = l1->val;
			l2val = l2->val;
			l1 = l1->next;
			l2 = l2->next;
			sum = carry + l1val + l2val;
			carry = sum/10;
			sum %= 10;
			cur->next = new ListNode(sum);
			cur = cur->next;
		}
		while (l1)
		{
			l1val = l1->val;
			l1 = l1->next;
			sum = carry + l1val;
			carry = sum/10;
			sum %= 10;
			cur->next = new ListNode(sum);
			cur = cur->next;
		}
		while (l2)
		{
			l1val = l2->val;
			l2 = l2->next;
			sum = carry + l1val;
			carry = sum/10;
			sum %= 10;
			cur->next = new ListNode(sum);
			cur = cur->next;
		}
		if (carry != 0)
		{
			cur->next = new ListNode(carry);
		}
		ListNode *t = sumList->next;
		delete sumList;
		return t;
	}

 
これは簡潔かもしれませんが、論理はあまり悪くありません.データの処理が少し違います.

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
		ListNode *sumList = new ListNode(-1);
		ListNode *cur = sumList;

		int carry = 0;
		while (l1 != nullptr || l2 != nullptr || carry != 0) 
		{
			int l1val = l1 == nullptr ? 0 : l1->val;
			int l2val = l2 == nullptr ? 0 : l2->val;
			int sum = l1val + l2val + carry;
			carry = sum / 10;
			sum %= 10;

			cur->next = new ListNode(sum);
			cur = cur->next;

			l1 = l1 == nullptr ? nullptr : l1->next;
			l2 = l2 == nullptr ? nullptr : l2->next;
		}

		ListNode *t = sumList->next;
		delete sumList;
		return t;
	}