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;    }};