2. Add Two Numbers

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本文介绍了一种算法，用于处理两个非空链表表示的非负整数相加的问题。链表中的数字以逆序存储，每个节点包含一个数字。通过遍历链表，逐位相加并处理进位，最终返回结果链表。

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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.

O(N) solution:(加法：从个位依次往上加，即从左到右)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode pre = new ListNode(-1);
        ListNode head = pre;
        int extral=0;
        int sum=0;
        while(l1!=null || l2!=null || extral!=0){
            sum = (l1 != null ? l1.val: 0) + (l2!=null ? l2.val : 0) + extral;
            ListNode one = new ListNode(sum%10);
            extral = sum / 10;
            pre.next = one;
            pre = one;
            if(l1!=null){
                l1 = l1.next;
            }
            if(l2!=null){
                l2 = l2.next;
            }
        }
        pre.next = null;
        return head.next ;
    }
}