Middle-题目110：2. Add Two Numbers

题目原文：
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
题目大意：
给出两个数，用链表逆序的存储每个数位，求他们的和，并也用链表逆序存储数位。
题目分析：
方法一：用BigInteger储存两个数字，再加起来，再生成对应链表。
方法二：模拟整数竖式算法（好在是逆序的， 正序就麻烦了）
源码：（language：java/c）
方法一：

import java.math.BigInteger;

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        BigInteger num1 = BigInteger.ZERO;
        BigInteger num2 = BigInteger.ZERO;
        BigInteger temp = BigInteger.ONE;

        while(l1!=null) {
            //num1+=temp*l1.val;
            num1=num1.add(temp.multiply(BigInteger.valueOf(l1.val)));
            temp=temp.multiply(BigInteger.TEN);
            l1=l1.next;
        }
        temp=BigInteger.ONE;
        while(l2!=null) {
            num2=num2.add(temp.multiply(BigInteger.valueOf(l2.val)));
            temp=temp.multiply(BigInteger.TEN);
            l2=l2.next;
        }
        BigInteger sum = num1.add(num2);
        if(sum.compareTo(BigInteger.ZERO)==0)
            return new ListNode(0);
        else {
            ListNode l = new ListNode(sum.mod(BigInteger.TEN).intValue());
            ListNode p=l;
            while(sum.compareTo(BigInteger.ZERO)!=0) {
                sum=sum.divide(BigInteger.TEN);
                if(sum.compareTo(BigInteger.ZERO)!=0) {
                    p.next = new ListNode(sum.mod(BigInteger.TEN).intValue());
                    p=p.next;
                }
            }
            return l;
        }
    }
}

方法二：(from discuss)

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode* result = l1 ? l1 : l2;
    struct ListNode* carry = l1 ? l2 : l1;
    struct ListNode* node = result;
    struct ListNode* tmp;
    int c = 0;
    int s;
    while (l1 || l2) {
        s = c;
        if (l1) { s += l1->val; l1=l1->next; }
        if (l2) { s += l2->val; l2=l2->next; }
        c = s > 9 ? 1 : 0;
        node->val = c ? s - 10 : s;
        if (l1) {
            node = node->next = l1;
        } else if (l2) {
            node = node->next = l2;
        } else {
            node->next = NULL;
        }
    }
    if (c) {
        carry->val = c;
        node->next = carry;
        node = node->next;
    }
    node->next = NULL;
    return result;
}

成绩：
方法一：28ms,beats 1.57%,众数4ms,63.50%
方法二：20ms,beats 18.86%,众数20ms,68.68%