2. Add Two Numbers

2. Add Two Numbers

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.

问题链接：https://leetcode.com/problems/add-two-numbers/
https://leetcode-cn.com/problems/add-two-numbers

解法说明

首先循环列表1和列表2相加，记录每次需要进位的值，不断让head.next指向产生的节点；
接着处理最长的列表剩下的节点
最后处理最长的节点或者列表1和列表2相加剩下的进位的值

整个解决方案如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int preSum = 0;
        int sum = 0;
        ListNode head = new ListNode(0);
        //记录头节点
        ListNode pre = head;
        //先直接计算两者，直到最短的退出
        while(l1 != null && l2 != null ){
            sum = l1.val+l2.val;
            sum += preSum;
            preSum = sum / 10;
            sum = sum % 10;
            ListNode currentNode = new ListNode(sum);
            head.next = currentNode;
            head =  currentNode;
            l1 = l1.next;
            l2 = l2.next;
        }

        ListNode restNode = (l1 == null) ? l2:l1;
        //短的结束后，另外一个继续计算和进位
        while(restNode != null){
            sum = restNode.val;
            sum += preSum;
            preSum = sum / 10;
            sum = sum % 10;
            ListNode temp = new ListNode(sum);
            head.next = temp;
            head = temp;
            restNode = restNode.next;
        }
        //最后还剩下进位的
        if(preSum > 0){
            ListNode lastNode = new ListNode(preSum);
            head.next = lastNode;
            head = lastNode;
        }
        return pre.next;
    }
}

官方的参考解决方案：
https://leetcode.com/problems/add-two-numbers/solution/