[LeetCode][2 Add Two Numbers][medium]Java实现

原题：
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.
示例：
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
题目的意思是用链表的方式实现两个非负整数的相加。每一位都是一个节点，且个位在最前面。简单点来说就是我们小学刚学数字加法的时候的进位。【大数的加法运算】

我最初的思路，一种递归的思想不停地计算下个节点的下个节点。。。

public class Solution2 {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        int x = l1.val + l2.val;        int i = 0;        if (x >= 10) {            x = x % 10;            i = 1;        }        ListNode listNode = new ListNode(x);        if (l1.next != null || l2.next != null || i != 0)            listNode.next = addTwoNumbersx(l1.next, l2.next, i);        return listNode;    }    private ListNode addTwoNumbersx(ListNode l1, ListNode l2, int j) {        int x;        int i = 0;        ListNode listNode;        int k1, k2;        ListNode next1;        ListNode next2;        if (l1 == null) {            k1 = 0;            next1 = null;        } else {            k1 = l1.val;            next1 = l1.next;        }        if (l2 == null) {            k2 = 0;            next2 = null;        } else {            k2 = l2.val;            next2 = l2.next;        }        x = k1 + k2 + j;        if (x >= 10) {            x = x % 10;            i = 1;        }        listNode = new ListNode(x);        if (next1 != null || next2 != null || i != 0)            listNode.next = addTwoNumbersx(next1, next2, i);        return listNode;    }}class ListNode {    int val;    ListNode next;    ListNode(int x) {        val = x;    }}

虽然Accepted了，但是效率并不高。后来看了另外一篇博客http://blog.csdn.net/DERRANTCM/article/details/46905467，觉得他的做法更加高效。

import java.util.List;public class Solution2_2 {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode p1 = l1;        ListNode p2 = l2;        ListNode root = new ListNode(0);        ListNode r = root;        int carry = 0;        int s;        while (p1 != null && p2 != null) {            s = p1.val + p2.val + carry;            carry = s / 10;            p1.val = s % 10;            r.next = p1;            r = p1;            p1 = p1.next;            p2 = p2.next;        }        if (p1 != null) {            r.next = p1;        } else if (p2 != null) {            r.next = p2;        }        if (carry == 1) {            while (r.next != null) {                s = r.next.val + carry;                r.next.val = s % 10;                carry = s / 10;                r = r.next;            }            if (carry > 0) {                r.next = new ListNode(1);            }        }        return root.next;    }    class ListNode {        int val;        ListNode next;        ListNode(int x) {            val = x;        }    }}

持续学习中ing。。。。