题解 | #链表相加(二)# Stack + 暴力

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<Integer> stk1 = new Stack<>();
        Stack<Integer> stk2 = new Stack<>();
        //Stack<Integer> result = new Stack<>();
        while(head1!=null){
            stk1.push(head1.val);
            head1 = head1.next;
        }
        while(head2!=null){
            stk2.push(head2.val);
            head2 = head2.next;
        }
        int maxLen = Math.max(stk1.size(), stk2.size());
        int result[] = new int[maxLen + 1];
        for(int i = 0; i < maxLen; i++){
            int a = stk1.isEmpty()?0:stk1.pop();
            int b = stk2.isEmpty()?0:stk2.pop();
            int sum = a + b + result[i];
            if(sum >= 10) {
                result[i] = sum - 10;
                result[i+1] = 1;
            }else{
                result[i] = sum;
            }

        }
        int hasPrefix = 0;
        ListNode vNode = new ListNode(-1);
        ListNode cur = new ListNode(-1);
        vNode.next = cur;
        for(int i=maxLen;i>=0;i--){
            System.out.print(result[i]);
            if(hasPrefix == 0 && result[i] == 0){
                continue;
            }else if(hasPrefix == 0 && result[i] != 0){
                cur.next = new ListNode(result[i]);
                cur = cur.next;
                hasPrefix = 1;
            }else{
                cur.next = new ListNode(result[i]);
                cur = cur.next;
            }
        }
        
        return vNode.next.next;
    }
}

善用数据结构的一些特性，尤其是对于链表，栈的使用是非常常见的。