题解 | 2025/7/28（链表）（Java）

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    public ListNode addInList (ListNode head1, ListNode head2) {
        // write code here
        head1 = reverse(head1);
        head2 = reverse(head2);
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;

        int flag = 0;   // 是否存在进位
        while (head1 != null || head2 != null) {
            flag += head1 == null ? 0 : head1.val;
            flag += head2 == null ? 0 : head2.val;
            if (head1 != null) head1 = head1.next;
            if (head2 != null) head2 = head2.next;
            head.next = new ListNode(flag % 10);
            head = head.next;
            flag /= 10;
        }
        if (flag != 0) {
            head.next = new ListNode(flag);
        }
        return reverse(dummy.next);
    }

    private ListNode reverse(ListNode head) {
        ListNode pre = null;
        for (ListNode temp; head != null; head = temp) {
            temp = head.next;
            head.next = pre;
            pre = head;
        }
        return pre;
    }
}