LeetCode 2 Add Two Numbers（链表操作）

题目来源：https://leetcode.com/problems/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

题目要求：给出两个链表l1和l2,将相对应的元素相加,如果产生进位则保留个位,相应的十位加在后面的对应元素上.

解体思路：

首先判断链表是否为空,若l1空,直接返回l2,相反返回l2

然后找到l1和l2链表最短的一个长度min_len,进入循环

循环过程中,为了减少空间的开销,可以将链表l1作为最终的结果链表,只需要改变l1的最大容量为l1+l2即可,另外相加过程中需要一个整型temp变量来保存进位.

循环结束之后,判断l1或l2空否(之前计算最短链表时可以做标记),然后将不空的链表中其他元素全部放入结果链表中即可.

提交代码：

 /**
  * 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) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
 //        ListNode *pResult = NULL;
 //        ListNode **pCur = &pResult;

         ListNode rootNode();
         ListNode *pCurNode = &rootNode;
         int a = ;
         while (l1 || l2)
         {
             int v1 = (l1 ? l1->val : );
             int v2 = (l2 ? l2->val : );
             int temp = v1 + v2 + a;
             a = temp / ;
             ListNode *pNode = new ListNode((temp % ));
             pCurNode->next = pNode;
             pCurNode = pNode;
             if (l1)
                 l1 = l1->next;
             if (l2)
                 l2 = l2->next;
         }
         if (a > )
         {
             ListNode *pNode = new ListNode(a);
             pCurNode->next = pNode;
         }
         return rootNode.next;
     }
 };

下面给出数组解决的代码,具体过程相同,只是数据存储结构不同：

 #include <bits/stdc++.h>
 #define MAX 1000010

 using namespace std;

 int main()
 {
     int n1,n2;
     while(~scanf("%d %d",&n1,&n2))
     {
         int *a=new int[n1+n2+];
         int *b=new int[n2];
         for(int i=;i<n1;i++)
             scanf("%d",&a[i]);
         for(int i=;i<n2;i++)
             scanf("%d",&b[i]);
         int minlen=min(n1,n2);
         int maxlen=n1+n2-minlen;
         int temp=;
         for(int i=;i<minlen;i++)
         {
             a[i]=a[i]+b[i]+temp;
             if(a[i]>=)
             {
                 temp=a[i]/;
                 a[i]=a[i]%;
             }
         }
         for(int j=minlen;j<maxlen;j++)
         {
                 if(maxlen==n1)
                     a[j]=a[j];
                 else
                     a[j]=b[j];
         }
         for(int i=;i<maxlen;i++)
             printf("%d ",a[i]);
         printf("\n");
     }
 }