Leetcode: Add Two Numbers (1)

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

Idea: This problem is the basic problem about linked list operation. Here I use a dummy head and delete the head before I return the head of the sum. Therefore, you do not need to check if the head is null when appending to the linked list.

/**
 * 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) {
        int carry = 0;
        ListNode *first = new ListNode(0);
        ListNode *head = first;
        ListNode *curr = first;
        while(l1 && l2) {
            int num = (l1->val + l2->val + carry)%10;
            carry =  (l1->val + l2->val + carry)/10;
            ListNode *a_node = new ListNode(num);
            curr->next = a_node;
            curr = curr->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1) {
            int num = (l1->val + carry)%10;
            carry = (l1->val + carry) / 10;
            ListNode *a_node = new ListNode(num);
            curr->next = a_node;
            curr = curr->next;
            l1 = l1->next;
        }
        while(l2) {
            int num = (l2->val + carry)%10;
            carry = (l2->val + carry) / 10;
            ListNode *a_node = new ListNode(num);
            curr->next = a_node;
            curr = curr->next;
            l2 = l2->next;
        }
        if(carry) {
            ListNode *a_node = new ListNode(carry);
            curr->next = a_node;
        }
        head = head->next;
        delete(first);
        return head;
    }
};