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) {
        // write code here
        // 解题思路：
        // 1、根据题意可知需要把2个链表相加,但由于高位在链表头部,所以需要对链表进行反转
        // 2、对每个节点相加，对10取余(%10)部分作为节点值,取整部分作为一下个节点相加的累加值
        // 3、结果计算完毕后，需要对链表再次进行反转
        ListNode rHead1 =  resever(head1 );
        ListNode rHead2 =  resever(head2 );
        int init = 0;
        ListNode root = new ListNode(-1);
        ListNode cur = root;

        while (rHead1 != null || rHead2 != null || init != 0) {
            int val1 = rHead1 == null ? 0 : rHead1.val ;
            int val2 = rHead2 == null ? 0 : rHead2.val ;
            int sum = init + val1 + val2;
            init = sum / 10;
            ListNode node = new ListNode(sum % 10);
            cur.next = node;
            cur = cur.next;
            if(rHead1 != null){
                rHead1 = rHead1.next;
            }

            if(rHead2 != null){
                rHead2 = rHead2.next;
            }
        }

        return resever(root.next);
    }

    private ListNode resever(ListNode head) {
        ListNode pre = null;
        ListNode next = null;

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

        return pre;
    }
}