ListNode preHead(0), *p = &preHead;
    int extra = 0;
    while (l1 || l2 || extra) {
        int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;
        extra = sum / 10;
        p->next = new ListNode(sum % 10);
        p = p->next;
        l1 = l1 ? l1->next : l1;
        l2 = l2 ? l2->next : l2;
    }
    return preHead.next;
 

leetCode演算法題目：Add Two Numbers   這是我見過的最簡潔的C++實現了，作為學習筆記整理出來，方便日後檢視

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

我的第一版：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
	ListNode *l3 = new ListNode();
	ListNode *l5;
	l5 = l3;
	int temp = 0;
	if (l1 != NULL)
	{
		l3->val = ((l1->val) + (l2->val)) % 10 + temp;   //將節點相加
		temp = ((l1->val) + (l2->val)) / 10;
		l3->next = (ListNode*)malloc(sizeof(struct ListNode));
		l3 = l3->next;
		l1 = l1->next;
		l2 = l2->next;
	}
	return l5;
}