leetcode 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.


把是否为null用一个int来表示，如果为null就标记为0，这样整个表达式会非常统一。

/**
 * 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) {
        ListNode t = new ListNode(0);
        ListNode cur = t;
        int carry = 0;
        ListNode p = l1;
        ListNode q = l2;
        while(p!=null || q!=null) {
            int x = (p!=null) ? p.val : 0; //第一次用这种格式，很简洁。
            int y = (q!=null) ? q.val : 0;
            int sum = carry + x + y; 
            carry = sum/10;
            cur.next = new ListNode(sum%10); 用%的方式将“是否需要进位”这个问题统一了起来，一律在此新建一个listnode
            cur = cur.next;
            if(p!=null){p = p.next;}
            if(q!=null){q = q.next;}
        }
        if(carry>0) {
            cur.next = new ListNode(carry);
        }
        return t.next;
    }
}