# 32ms C++ no reverse, no extra space, no change to original lists

• Just use a pointer to save the last non-9 digit which can take a possible carry generated by less significant bits.

``````class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* head = new ListNode(0);
ListNode* tail = head;
ListNode* cur = head; // use to store the last node which value is less than 9
ListNode* h1 = l1;
ListNode* h2 = l2;
// find the node which will be used to do alignment on addition
while (h1 && h2) {
h1 = h1->next;
h2 = h2->next;
}
ListNode* t; // the node as alignment marker
if (h1) { t = h1; h1 = l1; h2 = l2; } // set h1 to the longer one's head
else { t = h2; h1 = l2; h2 = l1; } // set h2 to the shorter one's head
while (h1) {
int val = h1->val;
h1 = h1->next;
if (t) t = t->next; // if t is not NULL, h1 is still ahead of h2
else { // t == NULL, should add h2's value
val += h2->val;
h2 = h2->next;
}
if (val >= 10) {
val %= 10;
cur->val++; // take the carry value, because the original val is less than 9, no impact to previous nodes
while (cur != tail) { // set all afterward nodes value to 0, e.g. 1(cur)->9->9(tail)->7 + 4 -> 2->0->0->1
cur = cur->next;
cur->val = 0;
}
}
ListNode* node = new ListNode(val);
tail->next = node;
tail = node;
if (tail->val < 9) cur = tail; // if the value is less than 9, it has the capacity to take an afterward carry
}

return head->val ? head : head->next;
}
};
``````

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