[LeetCode] 2. Add Two Numbers ☆☆

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

解法: 建立一个新链表，然后把输入的两个链表从头往后撸，每两个相加，添加一个新节点到新链表后面，注意要处理下进位问题。还有就是最高位的进位问题要最后特殊处理一下（这一步容易忽略！）。时间复杂度为O(n)，空间复杂度为O(n)。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(-1);  // 方便后续操作，最终返回result.next即可
        ListNode currNode = result;
        int carry = 0;  // 进位1，不进位0

        while ((l1 != null) || (l2 != null)) {
            int d1 = l1 == null ? 0 : l1.val;
            int d2 = l2 == null ? 0 : l2.val;
            int sum = d1 + d2 + carry;
            carry = sum > 9 ? 1 : 0;

            currNode.next = new ListNode(sum % 10);
            currNode = currNode.next;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }

        if (carry == 1) {
            currNode.next = new ListNode(1);
        }

        return result.next;
    }
}

ps: 不要采用先计算出两个链表的数后相加，最后将和用链表表示的算法。因为链表的长度可能很长，不一定能用int或double表示。