题解 | #链表相加(二)#

利用Stack FILO的特性 先将 链表的每个值Push进Stack后 每次Pop出值；
时间空间复杂度都为O(n)

using System;
using System.Collections.Generic;

/*
public class ListNode
{
	public int val;
	public ListNode next;

	public ListNode (int x)
	{
		val = x;
	}
}
*/

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<int> st1 = new Stack<int>();
        Stack<int> st2 = new Stack<int>();
        ListNode result = null;
        int jingwei = 0;

        while(head1 != null){
            st1.Push(head1.val);
            head1 = head1.next;
        }

        while(head2 != null){
            st2.Push(head2.val);
            head2 = head2.next;
        }

        while(st1.Count != 0 || st2.Count != 0 || jingwei != 0 ){
            int x = (st1.Count != 0) ? st1.Pop() : 0;
            int y = (st2.Count != 0) ? st2.Pop() : 0;

            int sum = x + y + jingwei;
            jingwei = sum/10;

            ListNode newNode = new ListNode(sum % 10);
            newNode.next = result;
            result = newNode;
        }
        return result;
    }
}