题解 | #链表相加(二)#

• 逆转两个链表
• 进行相加处理
• 再次逆转链表

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
        // 进行判空处理
        if (head1 == null)
            return head2;
        if (head2 == null) {
            return head1;
        }
        head1 = reverseList(head1);
        head2 = reverseList(head2);
        ListNode head = new ListNode(-1);
        ListNode nhead = head;
        int flag = 0;
        while(head1 != null || head2 !=null)
        {
            int val=0;
            if(head1 != null)
            {
                val += head1.val;
                head1 = head1.next;
            }
            if(head2 != null)
            {
                val += head2.val;
                head2= head2.next;
            }
            val += flag;
            if(val > 9)
            {
                val = val%10;
                flag = 1;
            }else{
                flag = 0;
            }
            nhead.next = new ListNode(val);
            nhead = nhead.next;
        }
        if(flag ==  1)
        {
            nhead.next = new ListNode(1);
        }
        return reverseList(head.next);

    }

    private ListNode reverseList(ListNode head) {
        if (head == null) return head;

        ListNode pre = null;
        ListNode cur = head;
        ListNode next;
        while (cur != null) {
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }

        return pre;
    }
}