445. Add Two Numbers II

Posted 2020-10-12 andywu

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
题目含义：做列表的加法（由地位开始）

 1     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
 2         if (l1 == null) return l2;
 3         if (l2 == null) return l1;
 4         Stack<Integer> stack1 = new Stack<>();
 5         Stack<Integer> stack2 = new Stack<>();
 6         while (l1 != null) {
 7             stack1.push(l1.val);
 8             l1 = l1.next;
 9         }
10         while (l2 != null) {
11             stack2.push(l2.val);
12             l2 = l2.next;
13         }
14 
15         int sum = 0;
16         ListNode list = new ListNode(0);
17         while (!stack1.isEmpty() || !stack2.isEmpty()) {
18             if (!stack1.isEmpty()) sum += stack1.pop();
19             if (!stack2.isEmpty()) sum += stack2.pop();
20             list.val = sum % 10;
21             ListNode head = new ListNode(sum / 10);
22             sum /= 10;
23             head.next = list;
24             list = head;
25         }
26         return list.val == 0 ? list.next : list;        
27     }