---
title: 'LeetCode 445. Add Two Numbers II'
disqus: hackmd
---

# LeetCode 445. Add Two Numbers II

## Description

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

## Example

Input: l1 = [7,2,4,3], l2 = [5,6,4]
Output: [7,8,0,7]

## Constraints

* The number of nodes in each linked list is in the range [1, 100].
* 0 <= Node.val <= 9
* It is guaranteed that the list represents a number that does not have leading zeros.

## Answer

此題做法和第2題類似，一樣是將list中每位數解析出來相加並考慮進位項，比較不同的是list排列並沒有反過來，為了我們要從個位數開始計算，所以我們將list先做reverse後再開始相加，最後要接新空間時也是往前接，如此順序才會是正確的。

struct ListNode* my_rever_list(struct ListNode *head);

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    int add = 0;
    l1 = my_rever_list(l1);
    l2 = my_rever_list(l2);
    struct ListNode *ans = (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode *tmp = NULL;
    ans->next = NULL;
    while(l1 != NULL || l2 != NULL){
        if(l1 != NULL && l2 != NULL){
            ans->val = (l1->val + l2->val + add) % 10;
            add = (l1->val + l2->val + add) / 10;
            l1 = l1->next;
            l2 = l2->next;
        }
        else if(l1 != NULL){
            ans->val = (l1->val + add) % 10;
            add = (l1->val + add) / 10;
            l1 = l1->next;
        }
        else{
            ans->val = (l2->val + add) % 10;
            add = (l2->val + add) / 10;
            l2 = l2->next;
        }
        if(add || l1 != NULL || l2 != NULL){
            tmp = (struct ListNode*)malloc(sizeof(struct ListNode));
            tmp->next = ans;
            ans = tmp;
        }
    }
    if(add){ans->val = 1;}
    return ans;
}

struct ListNode* my_rever_list(struct ListNode *head){
    struct ListNode *nxt = NULL, *cur = NULL;
    while(head != NULL){
        nxt = head->next;
        head->next = cur;
        cur = head;
        head = nxt;
    }
    return cur;
}

## Link
https://leetcode.com/problems/add-two-numbers-ii/
###### tags: `Leetcode`