链表求和 II

假定用一个链表表示两个数，其中每个节点仅包含一个数字。假设这两个数的数字顺序排列，请设计一种方法将两个数相加，并将其结果表现为链表的形式。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2
     */
    public ListNode addLists2(ListNode l1, ListNode l2) {
        // write your code here
        l1 = reverseList(l1);
        l2 = reverseList(l2);
        return addList(l1, l2);
    }
    private ListNode reverseList(ListNode listnode){
        ListNode head = null;
        ListNode node;
        if (listnode == null){
            return null;
        }
        while (listnode != null){
            node = new ListNode(listnode.val);
            node.next = head;
            head = node;
            listnode = listnode.next;
        }
        return head;
    }
    private ListNode addList(ListNode l1, ListNode l2){
        int val, carry = 0;
        ListNode head = null, node = null, l = null;
        while (l1 != null && l2 != null){
            val = l1.val + l2.val + carry;
            carry = val / 10;
            val = val % 10;
            node = new ListNode(val);
            node.next = head;
            head = node;
            l1 = l1.next;
            l2 = l2.next;
        }
        if (l1 != null){
           l = l1;
        }
        if (l2 != null){
           l = l2;
        }
        while (l != null){
            val = l.val + carry;
            carry = val / 10;
            val = val % 10;
            node = new ListNode(val);
            node.next = head;
            head = node;
            l = l.next;
        }
        if (carry != 0){
            node = new ListNode(carry);
            node.next = head;
            head = node;
        }
        return head;
    }
}