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

連結串列加減，不能反轉連結串列，只能藉助棧或者其他容器實現，

	public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		if (l1 == null)
			return l2;
		if (l2 == null)
			return l1;
		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;
		// head的下一個結點為curNode
		ListNode curNode = new ListNode(0);
		while (!s1.isEmpty() || !s2.isEmpty()) {
			if (!s1.isEmpty())
				sum += s1.pop();
			if (!s2.isEmpty())
				sum += s2.pop();
			// head.val儲存進位制位，head.val可能為0
			ListNode head = new ListNode(sum / 10);
			// curNode儲存結果
			curNode.val = sum % 10;
			head.next = curNode;
			// curNode往前移動，指向head
			curNode = head;
			// 此時sum儲存的是進位制位
			// 下次計算需要用到
			sum /= 10;
		}
		// 前導0的情況,
		// curNode為head的引用，可能為0
		if (curNode.val == 0)
			curNode = curNode.next;
		return curNode;
		// 前導0的情況
		// return curNode.val == 0 ?
		// curNode.next : curNode;
	}