Leetcode 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

考查的是連結串列的合併，類似多項式的合併，開始打算利用連結串列本身的空間去做，但考慮到兩個連結串列共享同一個空間，會導致斷鏈，故還是採取了增加空間複雜度

如果資料量小的話，可以先把連結串列轉換成整數，然後相交後再轉換成連結串列的形式

#include <iostream>
#include <algorithm>

using namespace std;

static int debug = 0;

struct ListNode{
    int val;
    ListNode *next;
    ListNode(int x):val(x),next(NULL){}
};

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2){
    if (l1 == NULL) return l2;
    if (l2 == NULL) return l1;
    ListNode *head = NULL;
    ListNode *l3 = head;
    
    int carry = 0;
    while (l1 != NULL && l2 != NULL) {
        int sum = l1->val + l2->val + carry;
        carry = sum/10;
        sum = sum%10;
        if (head == NULL) {
            head = new ListNode(sum);
            l3 = head;
        }else{
            ListNode *tmp = new ListNode(sum);
            l3 ->next = tmp;
            l3 = tmp;
        }
        l1 = l1->next;
        l2 = l2->next;
    }
    while (l1!=NULL) {
        int sum = carry + l1->val;
        carry = sum/10;
        sum = sum%10;
        ListNode *tmp = new ListNode(sum);
        l3 ->next = tmp;
        l3 = tmp;
        l1 = l1->next;
    }
    while (l2!=NULL) {
        int sum = carry + l2->val;
        carry = sum/10;
        sum = sum%10;
        ListNode *tmp = new ListNode(sum);
        l3 ->next = tmp;
        l3 = tmp;
        l2 = l2->next;
    }
    if (carry) {
        ListNode *tmp = new ListNode(carry);
        l3 ->next = tmp;
        l3 = tmp;
    }
    return head;
}


int main(){
    ListNode *a = new ListNode(5);
    ListNode *b = new ListNode(5);
    
    ListNode *c = addTwoNumbers(a,b);
    while (c != NULL) {
        cout<<c->val<<endl;
        c = c->next;
    }
}