面试题：Add Two Numbers（模拟单链表）

题干：

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

注：

题目中链表的定义为

 public class ListNode {
     int val;
     ListNode next;

     ListNode(int val) {
         this.val = val;
     }
 }

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分析：

　　题干中链表中的数是倒序存储的，题干中给的例子其实是：342+465=807。

　　我们定义一个节点dummyHead来指向结果的头结点，然后定义三个节点p,q,curr分别用来记录l1当前节点，l2当前节点，结果当前节点。

　　定义int carry来记录是否有进位。

public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummyHead = new ListNode(0);
        ListNode p = l1, q = l2, curr = dummyHead;
        int carry = 0;
        while (p != null || q != null) {
            int x = (p != null) ? p.val : 0;
            int y = (q != null) ? q.val : 0;
            int sum = carry + x + y;
            carry = sum / 10;
            curr.next = new ListNode(sum % 10);
            curr = curr.next;
            if (p != null)
                p = p.next;
            if (q != null)
                q = q.next;
        }
        if (carry > 0) {
            curr.next = new ListNode(carry);
        }
        return dummyHead.next;
    }