leetcode-【中等题】2. Add Two Numbers

题目

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

链接

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

答案

1、直接按顺序加就行，保存进位carry

2、到最后还需要考虑进位carry是否为0

代码

 /**
  * 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) {
         if(l1 == NULL)
         {
             return l2;
         }

         if(l2 == NULL)
         {
             return l1;
         }

         ListNode *ans = NULL;
         ListNode *point = NULL;
         int carry = ;
         int sum = ;
         while(l1 != NULL && l2 != NULL)
         {
             sum = carry + l1->val + l2->val;
             carry = sum / ;
             sum = sum % ;

             ListNode *value = new ListNode(sum);
             if(ans == NULL)
             {
                 ans = value;
             }

             if(point != NULL)
             {
                 point->next = value;
             }
             point = value;

             l1 = l1->next;
             l2 = l2->next;
         }

         while(l1 != NULL)
         {
             sum = carry + l1->val;
             carry = sum / ;
             sum = sum % ;
             ListNode *value = new ListNode(sum);
             point->next = value;
             point = value;

             l1 = l1->next;
         }

         while(l2 != NULL)
         {
             sum = carry + l2->val;
             carry = sum / ;
             sum = sum % ;
             ListNode *value = new ListNode(sum);
             point->next = value;
             point = value;

             l2 = l2->next;
         }

         if(carry != )
         {
             ListNode *value = new ListNode(carry);
             point->next = value;
             point = value;
         }

         return ans;
     }
 };