java - 将两个表示为链表的数字相加

Question

我正在尝试实现一个函数来添加两个表示为反向链接列表的数字。在我看来，代码是正确的，但在提供两个仅由一个数字组成的链表的输入时 [5] 。输出是 [0]，但它应该是 [0]->[1]

示例第一个列表: 5->6->3//代表数字 365 第二个列表: 8->4->2//代表数字 248 。结果列表:3->1->6//代表数字 613

有人可以告诉我我在逻辑上做错了什么吗？`

/**
 * 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) {
        //find length
        int carry = 0;
        ListNode sum = null;
        int x = 0;
        ListNode top = new ListNode(0);
        while(l1 != null || l2 != null){
            sum = new ListNode((l1.val+l2.val+carry)%10);
//            sum.val = (l1+l2+carry)%10;
            if(x==0){
                top.next = sum;
                x++;
            }
            carry = (l1.val+l2.val)/10;
            l1 = l1.next;
            l2 = l2.next;
            sum = sum.next;
        }
        if(l1 == null && l2 == null){
            //return top;

        }else if(l1 != null && l2 == null){
            while(l1 != null){
                sum = new ListNode((l1.val+carry)%10);
//            sum.val = (l1+carry)%10;
            if(x==0){
                top.next = sum;
                x++;
            }
            carry = (l1.val)/10;
            l1 = l1.next;
            sum = sum.next;
            }
            //return top;
        }else{
            while(l1 != null){
            sum = new ListNode((l2.val+carry)%10);
//            sum.val = (l2+carry)%10;
            if(x==0){
                top.next = sum;
                x++;
            }
            carry = (l2.val)/10;
            l2 = l2.next;
            sum = sum.next;
            }
            //return top;
        }
        if(carry == 1){
            sum = new ListNode(1);

            sum = sum.next;
        }
        return top.next;
    }
}`

Answer 1

这是一个解决方案(对 ListNode 使用“value”方法):

package so30544570;

public class Solution {
    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return ListNode.valueOf(l1.value() + l2.value());
    }

    public static void main(String[] args) {
        ListNode l1 = ListNode.valueOf(365);
        ListNode l2 = ListNode.valueOf(248);
        System.out.println("l1 = " + l1);
        System.out.println("l2 = " + l2);
        System.out.println("l1 + l2 = " + addTwoNumbers(l1, l2));
    }
}

和ListNode:

package so30544570;

public class ListNode {
    private int val;
    private ListNode next;

    private ListNode(int x) {
        val = x;
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder().append(val);
        if (next != null) {
            sb.append(" -> ").append(next.toString());
        }
        return sb.toString();
    }

    public int value() {
        int result = val;
        if (next != null) {
            result += 10 * next.value();
        }
        return result;
    }

    public static ListNode valueOf(final int i) {
        ListNode root = new ListNode(i % 10);
        if (i / 10 > 0) {
            root.next = ListNode.valueOf(i / 10);
        }
        return root;
    }
}

<小时/>

输出:

l1 = 5 -> 6 -> 3

l2 = 8 -> 4 -> 2

l1 + l2 = 3 -> 1 -> 6

<小时/>

注意:此解决方案避免求和计算，我们只需为求和重新创建一个ListNode。