2. Add Two Numbers

题目

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.

Example:

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

Explanation: 342 + 465 = 807.

题目大意

2 个逆序的链表，要求从低位开始相加，得出结果也逆序输出，返回值是逆序结果链表的头结点。

解题思路

需要注意的是各种进位问题。

极端情况，例如

Input: (9 -> 9 -> 9 -> 9 -> 9) + (1 -> )
Output: 0 -> 0 -> 0 -> 0 -> 0 -> 1

为了处理方法统一，可以先建立一个虚拟头结点，这个虚拟头结点的 Next 指向真正的 head，这样 head 不需要单独处理，直接 while 循环即可。另外判断循环终止的条件不用是 p.Next ！= nil，这样最后一位还需要额外计算，循环终止条件应该是 p != nil。

代码解析

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode cur = head;
        int carry = 0;
        while (l1 != null || l2 != null || carry != 0){
            int val = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = val / 10;
            cur.next = new ListNode(val % 10);
            cur = cur.next;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        return head.next;
    }
}