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) {
        if (head1 == null) {
            return head2;
        }
        if (head2 == null) {
            return head1;
        }

        // 将链表进行反转，然后相加，最后将相加的结果进行反转
        ListNode node1 = reverse(head1);
        ListNode node2 = reverse(head2);
        int carry = 0; // 进制
        ListNode pre = new ListNode(-1);
        ListNode head = pre;
        while (node1 != null || node2 != null || carry != 0) {
            int val1 = node1 == null ? 0 : node1.val;
            int val2 = node2 == null ? 0 : node2.val;

            int sum = val1 + val2 + carry;
            carry = sum / 10;
            sum = sum % 10;
            head.next = new ListNode(sum);
            head = head.next;

            if (node1 != null) {
                node1 = node1.next;
            }
            if (node2 != null) {
                node2 = node2.next;
            }
        }
        return reverse(pre.next);
    }

    public ListNode reverse(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode cur = head;
        ListNode pre = null;
        while (cur != null) {
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
}

代码的截图思路和刚才两个数相加的思路是一致的，不过是通过链表进行操作而不是数组或者字符串