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。

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