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

這道題可以看作兩個鏈表的相加，也可以看成342+465=807，只不過換作鏈表的形式。

/**
 * 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) {
       int carry =0;
 
        ListNode newHead = new ListNode(0);
        ListNode p1 = l1, p2 = l2, p3=newHead;
 
        while(p1 != null || p2 != null){//兩鏈表數(shù)據(jù)相應的節(jié)點相加，carry用于儲存之前進位的數(shù)據(jù)
            if(p1 != null){
                carry += p1.val;
                p1 = p1.next;
            }
 
            if(p2 != null){
                carry += p2.val;
                p2 = p2.next;
            }
 
            p3.next = new ListNode(carry%10);//將運算結果儲存在newHead的next節(jié)點處
            p3 = p3.next;
            carry /= 10;//進位的值
        }
 
        if(carry==1) 
            p3.next=new ListNode(1);//最后判斷最高位是否進一位，如果大于9，就再創(chuàng)建一個節(jié)點。
 
        return newHead.next;
    }
}