LeetCode-Add Two Numbers II

Description

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

Solution

思路：
这里使用了两个栈将两个链表的值反置；然后一次按位相加即可得到和。
这里有几个点需要注意下：
1. 相加后得到的数据是从低位开始的。需要进行反置(或者直接在返回值上做文章)
2. 按位相加得到和后，需要注意下最后是否还需要高位进一；如果需要则需要补充一个高位。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        Stack<Integer> sta1 = new Stack<Integer>();        Stack<Integer> sta2 = new Stack<Integer>();        while (l1 != null) {            sta1.push(l1.val);            l1 = l1.next;        }        while (l2 != null) {            sta2.push(l2.val);            l2 = l2.next;        }        ListNode re  = null;        int x = 0;        while (!sta1.isEmpty() || !sta2.isEmpty()) {            x += (sta1.isEmpty() ? 0 : sta1.pop()) + (sta2.isEmpty() ? 0 : sta2.pop());            ListNode node = new ListNode(x%10);            node.next = re;            re = node;            x = x / 10;        }        if(x ==1){            ListNode node = new ListNode(1);            node.next = re;            re = node;        }        return re;    }}