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
        Stack<ListNode>  st1 = new Stack<>();
        Stack<ListNode> st2 = new Stack<>();
        ListNode start = new ListNode(-1);
        while (head1 != null) {
            st1.add(head1);
            head1 = head1.next;
        }
        while (head2 != null) {
            st2.add(head2);
            head2 = head2.next;
        }
        int jinwei = 0;
        while (!st1.isEmpty() && !st2.isEmpty()) {
            int nodeVal = st1.pop().val + st2.pop().val + jinwei;
            jinwei = nodeVal / 10;
            ListNode temp = new ListNode(nodeVal % 10);
            temp.next = start.next;
            start.next = temp;
        }
        while (!st1.isEmpty()) {
            int nodeVal = st1.pop().val + jinwei;
            jinwei = nodeVal / 10;
            ListNode temp = new ListNode(nodeVal % 10);
            temp.next = start.next;
            start.next = temp;
        }
        while (!st2.isEmpty()) {
            int nodeVal = st2.pop().val + jinwei;
            jinwei = nodeVal / 10;
            ListNode temp = new ListNode(nodeVal % 10);
            temp.next = start.next;
            start.next = temp;

        }
        return start.next;
    }
}

我是利用了栈。如果没其他性能要求的话，这样还是最稳妥的。