题解 | #链表相加(二)#

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) {
        // 反转链表，使个位数在前
        ListNode num1 = reverse(head1);
        ListNode num2 = reverse(head2);

        // 头节点之前再创建一个节点，方便处理
        ListNode hair = new ListNode(0);
        ListNode pointer = hair;

        // 进位数
        int res = 0;

        // 从个位开始相加
        while (num1 != null || num2 != null) {
            int sum = 0;
            if (num1 == null) {
                sum = res + num2.val;
                num2 = num2.next;
            } else if (num2 == null) {
                sum = res + num1.val;
                num1 = num1.next;
            } else {
                sum = res + num1.val + num2.val;
                num1 = num1.next;
                num2 = num2.next;
            }
            int val = sum % 10;
            res = sum / 10;
            ListNode current = new ListNode(val);
            pointer.next = current;
            pointer = pointer.next;
        }

        // 处理最后一个进位
        if (res > 0) {
            pointer.next = new ListNode(res);
        }

        // 反转链表
        return reverse(hair.next);
    }

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