Leetcode: Add two numbers: complete project codes in C++ with detail comments

The key points in this problem:

1. linked list
2. how to build a linked listed in run time
For the second problem, you need to understand the following two line of codes:

//create new node in run time and change the pointer in tail

p_result_tmp->next = new ListNode(tmp);

p_result_tmp = p_result_tmp->next;

As in list struct,
  p_result_tmp->val   is the value 
 p_result_tmp->next  is the pointer to the linked list
 p_result_tmp        is the pointer/address to itself 
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

//////////////////////////////////////////////////////////////////

//Add Two Numbers

//////////////////////////////////////////////////////////////////

# include "stdafx.h"

# include <cstdlib>

# include <iostream>     // std::cout

using namespace std;

/////////////////////////////////////////////

//Add Two Numbers

/////////////////////////////////////////////

//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* p1;

              ListNode* p2;

              ListNode* p_result;

              p_result = new ListNode(0);

              ListNode* p_result_tmp;

              p_result_tmp = p_result;

              int add_tmp=0;

              int tmp = 0;

              int c=0;

              p1 = l1;

              p2 = l2;

              // using while loop as the length of list is not known ahead

              while (p1!=NULL || p2!=NULL){

           

                     if (p1 != NULL && p2 != NULL){

                            add_tmp = p1->val + p2->val+c;

                           c = add_tmp > 9 ? 1 : 0;

                           tmp = c == 0 ? add_tmp : add_tmp % 10;

                           //creat new node in run time and change the pointer in tail

                           p_result_tmp->next = new ListNode(tmp);

                           p_result_tmp = p_result_tmp->next;

                           //pointer control

                           p1 = p1->next;

                           p2 = p2->next;

                     }

                     else{

                           tmp = p1 != NULL ? p1->val + c : p2->val + c;

                           //check whether tmp >=10

                           c = tmp > 9 ? 1 : 0;

                           tmp = c == 0 ? tmp : tmp % 10;//take second digit

                           p_result_tmp->next = new ListNode(tmp);

                           p_result_tmp = p_result_tmp->next;

                           //pointer control

                           if (p1 != NULL){

                                  p1 = p1->next;

                           }

                           else{

                                  p2 = p2->next;

                           }

                     }

              }

              //the condition when input is like [5], [5]

              if (c == 1){

                     p_result_tmp->next = new ListNode(1);

                     p_result_tmp = p_result_tmp->next;

              }

              p_result = p_result->next;//exclude the first one

              return p_result;

       }

};

int main(int argc, char *argv[])

{

       ListNode* l1; //define a pointer that points to struct

       l1 = new ListNode(9);//assign a value to the pointer

       l1->next = new ListNode(9);//assign a value to the next linked pointer

       ListNode* l2;

       l2 = new ListNode(1);// why need new??

       l2->next = new ListNode(9);

       l2->next->next = new ListNode(8);

       Solution s;

       ListNode* result;

       result = s.addTwoNumbers(l1,l2);

      

       cout << "the solution is";

       cout << result->val;

       cout << "  Hello, world!\n";

       int age;

       cin >> age;

       return 0;

}