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
这题虽然归类到Medium 但是难度不大，只是要考虑的东西有点多，比如：两个list不一定一样长，进位这两个问题。
因为两个数字是逆序存到链表里的，所以这大大简化了问题难度，因为只需要向着链表遍历方向往前进位就可以了。
void addIter(ListNode *l1, ListNode *l2, ListNode *prev, int carry) {
if (!l1 && !l2 && carry == ) return;
ListNode *node;
int val = ;
if (l1) {
val += l1->val;
}
if (l2) {
val += l2->val;
}
val += carry;
carry = val / ;
val = val % ;
node = new ListNode(val);
if (prev) {
prev->next = node;
}
addIter(l1 ? l1->next : NULL, l2 ? l2->next : NULL, node, carry);
}
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode *ret = new ListNode();
addIter(l1, l2, ret, );
return ret->next;
}