2. Add Two Numbers

Description

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
Explanation: 342 + 465 = 807.

Solution

简单的链表加法，说实话，定义成medium有点牵强。注意进位信息，while的判定条件很妙。在做链表题的时候，经常会遇到首节点不方便处理，注意虚拟节点dummy的用法，能省去不少边界条件的判定。
时间O(m+n)，空间O(m+n)，能够AC，没有考虑去复用l1或者l2的空间

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    auto dummy = new ListNode(-1), pNode = dummy;
    int carry = 0;
    while (l1 || l2 || carry) {
        int sum = (l1?l1->val:0) + (l2?l2->val:0) + carry;
        pNode->next = new ListNode(sum%10);
        carry = sum/10;
        pNode = pNode->next;
        l1 = (l1?l1->next:NULL);
        l2 = (l2?l2->next:NULL);
    }
    return dummy->next;
}