Programming Skills: Add Two Numbers

[題目]

https://leetcode.com/problems/add-two-numbers/

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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

出處 leetcode https://leetcode.com/problems/add-two-numbers/description/

兩個數字，分別用link list來表示，請把他們相加後，用link list回傳結果。
位數在list中順序是顛倒的
e.g.
數字342 表示為 2 -> 4 -> 3

[思路]
數字相加的方法就是從低位開始，同位數相加再加上之前的進位，例如十位數相加再加上由個位數來的進位值。加出來的sum%10為本位數的結果，sum/10為進位值。

知道這個原理，那麼就用while loop同時爬兩個list，每個loop round都處理一個位數的相加。每個位數算出來後都建立一個新的node，把node插入result list中。


/**
 * 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 d(-1);//dummy head of result list
        d.next=NULL;
        ListNode* cur = &d;
        int carry=0;
        
        //main part
        while(l1 || l2 || carry!=0){
            int d1 = l1 ? l1->val : 0;
            int d2 = l2 ? l2->val : 0;
            int sum=d1+d2+carry;
            carry = sum/10;
            cur->next = new ListNode(sum%10);
            cur = cur->next;
            l1 = l1 ? l1->next : NULL;
            l2 = l2 ? l2->next : NULL;
        }
        
        return d.next;
    }
};