LeetCode解析（二）：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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

难度：Medium

解析：显然这个需要同步的将两个链表进行对应节点的循环相加，直至两个链表均无节点，且无进位，否则继续进行循环。不过，在循环相加的时候需要注意各种异常情况，如两链表长短不一，或者进位问题，导致循环结束的时机不同。

方法一：穷举遍历（★★★）

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode p = new ListNode(0);
        ListNode result = p;
        int isAdd = 0;
        while(l1 != null || l2 != null || isAdd != 0){
            int temp = (l1 != null ? l1.val : 0) + (l2 != null ? l2.val : 0) + isAdd;
            
            ListNode node = new ListNode(temp%10);
            p.next = node;
            p = node;
            isAdd = temp/10; 
            
            if(l1 != null){
                l1 = l1.next;
            }
            if(l2 != null){
                l2 = l2.next;
            }  
        }
        return result.next;
    }
}

结果：Runtime:42 ms，beats 20.69% of Java submission.
备注：遍历两个链表，直至两个量表均无节点且无进位，时间复杂度O(n)。