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445. Add Two Numbers 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
題目含義:做列表的加法(由地位開始)

 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 }

445. Add Two Numbers II