【LeetCode】Add Two Numbers

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

 

按照定义做，逐位相加，注意进位。

/**
 * 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* newhead = new ListNode(-1);
        ListNode* tail = newhead;
        int carry = 0;
        int sum;
        int val;
        while(l1 != NULL && l2 != NULL)
        {
            sum = l1->val + l2->val + carry;
            val = sum%10;
            carry = sum/10;
            tail->next = new ListNode(val);
            tail = tail->next;
            
            l1 = l1->next;
            l2 = l2->next;
        }
        if(l1 == NULL)
        {
            if(l2 == NULL)
            {//l1, l2 == NULL
                if(carry != 0)
                    tail->next = new ListNode(carry);
            }
            else
            {//l1 == NULL, l2 != NULL
                while(l2 != NULL)
                {
                    sum = l2->val + carry;
                    val = sum%10;
                    carry = sum/10;
                    tail->next = new ListNode(val);
                    tail = tail->next;
                    l2 = l2->next;
                }
                if(carry != 0)
                    tail->next = new ListNode(carry);
            }
        }
        else
        {//l1 != NULL, that means l2 == NULL
            while(l1 != NULL)
            {
                sum = l1->val + carry;
                val = sum%10;
                carry = sum/10;
                tail->next = new ListNode(val);
                tail = tail->next;
                l1 = l1->next;
            }
            if(carry != 0)
                tail->next = new ListNode(carry);
        }
        return newhead->next;
    }
};