Leetcode:Add Two Numbers

题目：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

分析：模拟两个数的加法，只不过这两个数是用链表（l1,l2）表示的。这里注意的是进位，我们用一个变量offset来表示该位上是否有来自低位的进位，初始offset = 0.
取l1,l2的头结点进行加法操作（还要加上offset）得到结果sum，如果sum<10，新建一个结点，它的值为sum,offset = 0;如果sum>=10,新建结点的值为sum-10,同时offset = 1。直到有一个链表到达表尾。然后我们考虑另一条链表。最后两个链表都遍历完了之后，我们还需查看offset是否=1，如果是，我们需再链上一个新的结点，它的值为1。

代码如下：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        if(l1 == null && l2 == null)            return null;        ListNode h = new ListNode(0);        h.next = null;        ListNode pre = h;        int offset = 0;        while(l1 != null && l2 != null)        {            int sum = l1.val+l2.val+offset;            if(sum > 9)            {                sum -= 10;                offset = 1;            }            else                offset = 0;            ListNode tmp = new ListNode(sum);            pre.next = tmp;            pre = tmp;            l1 = l1.next;            l2 = l2.next;        }        while(l1 != null)        {            int sum = l1.val+offset;            if(sum > 9)            {                sum -= 10;                offset = 1;            }            else                offset = 0;            ListNode tmp = new ListNode(sum);            pre.next = tmp;            pre = tmp;            l1 = l1.next;        }        while(l2 != null)        {            int sum = l2.val+offset;            if(sum > 9)            {                sum -= 10;                offset = 1;            }            else                offset = 0;            ListNode tmp = new ListNode(sum);            pre.next = tmp;            pre = tmp;            l2 = l2.next;        }        if(offset == 1)        {            ListNode t = new ListNode(1);            t.next = null;            pre.next = t;        }        return h.next;    }