[leetcode]445. Add Two Numbers II 两数相加II

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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

题目

俩数用链表表示，做个加法

思路

1. use stack to covert input list data

2. sum up from pop item from two stacks to a desired linkedlist

代码

 class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
         if (l1 == null && l2 == null) return null;
         if (l2 == null) return l1;
         if (l1 == null) return l2;
         // use stack to convert list's order
         Stack<Integer> s1 = new Stack<>();
         Stack<Integer> s2 = new Stack<>();

         while (l1 != null) {
             s1.push(l1.val);
             l1 = l1.next;
         }

         while (l2 != null) {
             s2.push(l2.val);
             l2 = l2.next;
         }

         int sum = 0;
         ListNode head = new ListNode(0);
         // sum up to a new linkedlist
         while (!s1.isEmpty() || !s2.isEmpty()) {
             if (!s1.isEmpty()) sum += s1.pop();
             if (!s2.isEmpty()) sum += s2.pop();

             int val = sum % 10;

             ListNode node = new ListNode(0);
             head.val = val;
             node.next = head;
             head = node;

             sum /= 10;
         }
         if (sum == 0){
             return head.next;
         }
         head.val = sum;
         return head;
     }
 }