LeetCode算法题2：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

 public class LeeAddTwoNum{    public static void main(String[] args)    {        // TODO Auto-generated method stub        ListNode s, r,s2,r2;        int x= 0;        LeeAddTwoNum latn = new LeeAddTwoNum();        ListNode Head1 = new ListNode(0);        Scanner scanner = new Scanner(System.in);        //链表首部，单独处理        r = Head1;        x = scanner.nextInt();        r.val = x;        //接收输入，以29结束        x = scanner.nextInt();        while (x != 29)        {            s = new ListNode(x);            r.next = s;            r = s;            x = scanner.nextInt();        }        System.out.println();        //第二个        ListNode Head2 = new ListNode(0);        r2 = Head2;        x = scanner.nextInt();        r2.val = x;        x = scanner.nextInt();        while (x != 29)        {            s2 = new ListNode(x);            r2.next = s2;            r2 = s2;            x = scanner.nextInt();        }        ListNode Ln;        Ln = Head1;        while(Ln!=null)        {            System.out.print(Ln.val+"-->");            Ln = Ln.next;        }        System.out.println("LL");        Ln = Head2;        while(Ln!=null)        {            System.out.print(Ln.val+"-->");            Ln = Ln.next;        }        System.out.print("end");        System.out.println();        latn.addTwoNumbers(Head1, Head2);    }    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode r,s,Head;        int flag=0,i1,i2,i;        Head = new ListNode((l1.val+l2.val)%10);        r=Head;        flag=(l1.val+l2.val)/10;        l1=l1.next;        l2=l2.next;        while((l1!=null)||(l2!=null))        {            if(l1==null)            {                i1=0;            }            else            {                i1=l1.val;                l1=l1.next;            }        /////////////////////////////               if(l2==null)            {                i2=0;            }            else            {                i2=l2.val;                l2=l2.next;            }            i=(i1+i2+flag)%10;            flag=(i1+i2+flag)/10;            ////////////////////             s = new ListNode(i);             r.next = s;             r=s;        }        if(flag!=0)        {             r.next = new ListNode(flag);        } //打印链表      //     while(Head!=null)    //  {    //      System.out.print(Head.val+"-->");    //      Head = Head.next;    //  }    //  System.out.print("end");        return Head;    }    //可以处理多位数的情况    public ListNode addTwoNumbers1(ListNode l1, ListNode l2) {        ListNode r,s,Head;        long i1=0;        long i2=0;        //取得链表里的数据，转为整数        StringBuffer sb =new StringBuffer();        StringBuffer sb2 =new StringBuffer();        while(l1!=null)        {            sb.append(l1.val);            l1=l1.next;        }        while(l2!=null)        {            sb2.append(l2.val);            l2=l2.next;        }     i1=Long.valueOf(sb.reverse().toString());   i2=Long.valueOf(sb2.reverse().toString());        i1=i1+i2;        System.out.println("add="+i1);        StringBuffer res = new StringBuffer(String.valueOf(i1)).reverse();        System.out.println("rese="+res.toString());        System.out.println();        int i;        Head = new ListNode(Integer.valueOf(String.valueOf(res.charAt(0))));        r=Head;        for(i=1;i<res.length();i++)        {            s = new ListNode(Integer.valueOf(String.valueOf(res.charAt(i))));             r.next = s;             r=s;        }        while(Head!=null)        {            System.out.print(Head.val+"-->");            Head = Head.next;        }        System.out.print("end");        return Head;    }    public static class  ListNode    {        int val;        ListNode next;        ListNode(int x)        {            val = x;        }    }}

这种做法的复杂度有点高，希望有更好的方案。