Leetcode 第二题，两数链表相加

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问题：

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

即，两个整数链表，相加，返回一个整数链表，注意进位。

解法：其实这个题名并没有捷径和太多技巧，主要是看是否理解链表的结构。

如果学过C/C++的指针的话，理解起来会更加容易。

 /** * Created by Atlantis on 16/4/25. *ListNode.java * Definition for singly-linked list. */public class ListNode {        int val;        ListNode next;        ListNode(int x){            val = x;        }}/** *Solution.java    * Created by Atlantis on 16/4/25. */public class Solution {    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {        if(null == l1 && null == l2){            return  null;        }        int forward = 0;        ListNode ret = new ListNode(0);        ListNode head = ret;        while (null != l1 && null != l2){            int sum = l1.val + l2.val +forward;            forward = sum/10;            l1 = l1.next;            l2 = l2.next;            ret.next = new ListNode(sum%10);            ret = ret.next;        }        while (null != l1){            int sum = l1.val+forward;            ret.next = new ListNode(sum%10);            forward = sum/10;            ret = ret.next;            l1 = l1.next;        }        while (null != l2){            int sum = l2.val+forward;            ret.next = new ListNode(sum%10);            forward = sum/10;            ret = ret.next;            l2 = l2.next;        }        if(forward != 0 ){            ret.next = new ListNode(forward);        }        return head.next;    }    public static void main(String[] args){        ListNode l1 = new ListNode(6);        l1.next = new ListNode(4);        l1.next.next = new ListNode(3);        ListNode l2 = new ListNode(5);        l2.next = new ListNode(6);        l2.next.next = new ListNode(4);        ListNode ret =addTwoNumbers(l1,l2);        if(null != ret) {            System.out.print(ret.val);        }        while (ret.next != null){            System.out.print("->");            System.out.print(ret.next.val);            ret = ret.next;        }    }}

结束：难度不是很大，会一点C/C++，比起双向链表什么的，有点无聊。。。

 链表这个东西，果然还是指针理解起来比较容易和直观。

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