Leet Code OJ 2. Add Two Numbers [Difficulty: Medium]

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

翻译：
给你2个链表，代表2个非负整数。链表中整数的每一位数字的存储是反序的，数组的每个节点都包含一个数字。把2个非负整数相加，并且用一个链表返回。
输入: (2 -> 4 -> 3) + (5 -> 6 -> 4)
输出: 7 -> 0 -> 8

分析：
由于刚好链表中数的存储是反序，也就是个位在最前面，正好方便我们从低位开始加。需要注意的几个点：
1. 传入[][],也就是2个空链表，要返回null
2. 传入[0][0]，也就是2个整数是0的处理，要返回[0]
3. 传入[5][5],要新增一个节点，存储进位。也就是判断是否结束，要根据2个链表是否为空和是否有进位来判断。

Java版代码：

/*** 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;}ListNode head=new ListNode(0);ListNode currentNode=head;int fix=0;while(l1!=null||l2!=null||fix!=0){int val1=0;int val2=0;if(l1!=null){val1=l1.val;l1=l1.next;}if(l2!=null){val2=l2.val;l2=l2.next;}int sum=val1+val2+fix;fix=0;if(sum>9){fix=1;sum-=10;}currentNode.next=new ListNode(0);currentNode.next.val=sum;currentNode=currentNode.next;}if(head.next==null){head.next=new ListNode(0);}return head.next;}
}

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