445. Add Two Numbers II

445. Add Two Numbers II

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

My Solutions:

• 如果没有不可以反转的限制条件，可以反转两个list，变成个位在最前，然后和题目#2解法相同。

• 不可以反转，则用stack，让个位加在stack最上

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> s1 = new Stack<>(), 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 dummy = new ListNode(-1);
        while (!s1.isEmpty() || !s2.isEmpty()) {
            if (!s1.isEmpty()) sum += s1.pop();
            if (!s2.isEmpty()) sum += s2.pop();
            
            dummy.val = sum % 10;
            ListNode head = new ListNode(sum / 10);
            
            head.next = dummy;
            dummy = head;
            sum = sum / 10;
        }
        
        return dummy.val == 0? dummy.next : dummy;
    }
}