Add Two Numbers

题目

Total Accepted: 137297　　Total Submissions: 599150　　Difficulty: Medium

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

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解题思路

任意选择一个Node进行迭代，迭代中有几种可能 1、Node1个数少于Node2 2、Node1个数和Node2相等 3、Node1个数大于Node2 分别进行考虑，计算进位和取模，迭代过程中改变Node1及Node2的偏移量，然后从当前偏移量Node2开始Node2的迭代。

代码

    /*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/public class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {if(l1==null||l2==null)return null;int multiple=0;ListNode result=null,nextNode=null;//用来表示结果与下一个Nodewhile(l1!=null){int n1=0,n2=0,sum=0;n1= l1.val;if(l2!=null){n2= l2.val;sum=n1+n2;sum+=multiple;multiple=0;if(sum>9){multiple=sum/10;sum=sum%10;}if(result==null){result=new ListNode(sum);nextNode=result;}else{ nextNode.next=new ListNode(sum);nextNode=nextNode.next;}l2=l2.next;//l2的指针随l1偏移l1=l1.next;}else{n1+=multiple;multiple=0;if(n1>9){multiple=n1/10;n1=n1%10;}nextNode.next=new ListNode(n1);nextNode=nextNode.next;l1=l1.next;}}while(l2!=null){//如果l1链表长度少于l2链表长度int n2= l2.val;n2+=multiple;multiple=0;if(n2>9){multiple=n2/10;n2=n2%10;}nextNode.next=new ListNode(n2);nextNode=nextNode.next;l2=l2.next;}if(multiple!=0){//迭代完Node1、Node2 最后如果进位不为0 则加一位nextNode.next=new ListNode(multiple);nextNode=nextNode.next;}return result;}}

复杂度

O(n)的时间 O(n)的内存

测试结果

其他思路

后面看到有用递归做这个题目的思路，代码会精简很多，大家可以尝试下

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