【Leetcode】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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode *p1 = l1, *p2 = l2, dummy(-1), *tail = &dummy;
        int sum = 0, carry = 0;
        while (p1 != nullptr || p2 != nullptr) {
            int v1 = p1 == nullptr ? 0 : p1->val;
            int v2 = p2 == nullptr ? 0 : p2->val;
            p1 = p1 == nullptr ? nullptr : p1->next;
            p2 = p2 == nullptr ? nullptr : p2->next;

            sum = v1 + v2 + carry;
            tail->next = new ListNode(sum % 10);
            tail = tail->next;
            carry = sum / 10;
        }
        if (carry > 0) {
            tail->next = new ListNode(carry);
            tail = tail->next;
        }
        return dummy.next;
    }
};

与Add binary类似，思想与方法完全一样，只是用链表实现，在链表前加哑元素可以使尾插入的代码简化。