Leetcode: Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

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

这道题挺常见的，关键是要设计出好的程序结构，有一些技巧，比如建立一个假的前置节点ListNode prenode = new ListNode(-1); 比如循环条件设置为 while (l1 != null || l2 != null || carry != 0)，我第一遍写的时候就是没有想到用“或”来写循环结构，导致下面有好多种子情况需要一一讨论，使得程序变得非常繁复，不似这样写架构清晰，思路明确。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null && l2 != null) return l2;
        if (l1 != null && l2 == null) return l1;
        if (l1 == null && l2 == null) return null;
        
        ListNode prenode = new ListNode(-1);
        ListNode end = prenode;
        int carry = 0;
        int current = 0;
        while (l1 != null || l2 != null || carry != 0) {
            current = 0;
            if (l1 != null) {
                current += l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                current += l2.val;
                l2 = l2.next;
            }
            if (carry != 0) {
                current += carry;
            }
            end.next = new ListNode(current % 10);
            carry = current / 10;
            end = end.next;
        }
        return prenode.next;
    }
}