Description

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.

Solution:

/**
* 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 temp1 = l1;
        ListNode temp2 = l2;
        ListNode res = null;
        ListNode fin = null;
        int prev = 0;
        while(temp1 != null || temp2 != null || prev != 0) {
            int temp = (temp1 == null ? 0 : temp1.val) + (temp2 == null ? 0 : temp2.val);
            temp = temp + prev;
            ListNode p = null;
            if (temp >= 10) {
                p = new ListNode(temp - 10);
                prev = 1;
            } else {
                p = new ListNode(temp);
                prev = 0;
            }
            if (res == null) {
                res = p;
                fin = p;
            } else {
                res.next = p;
            }
            res = p;
            temp1 = temp1 == null ? null : temp1.next;
            temp2 = temp2 == null ? null : temp2.next;
        }
        return fin;
    }
}

Attentions

这道题思路很简单，其实就是数学中数字相加的基本原理，链表从左向右遍历，实际上就是从低位遍历到高位，小学数学，低位相加和大于等于10，需要向上进一位，考虑到两个链表长度可能不一致，需要进行判空

SubMissions

Runtime:20 ms, faster than 82.57% of Java online submissions for Add Two Numbers.
Memory Usage:48.3 MB, less than 9.84% of Java online submissions for Add Two Numbers.
说明 速度仍然有优化的空间，同时内存占用稍高
重新分析占用：
思考使用递归的方法进行求和

submission.png