LeetCode|Add Two Numbers

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题目

You are given two linked lists representing two non-negative numbers. 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.

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

题目大意：给你两个链表，链表中没有负数，数字式翻转进入链表的，如：(2->4->3)则是342，链表的每个节点只能是一位的数组，需要得出两个链表相加后的一个链表。
思路：这个题目不难，需要考虑全面。1. 给的两个链表是否为空。2. 两个链表的长度有可能不相同。3. 最后一次相加是否需超过10。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        if (l1 == null) return l2;        else if (l2 == null) return l1;        ListNode currentNode = new ListNode(0);        ListNode head = currentNode;        int isNeedPlusOne = 0;        while (l1 != null && l2 != null) {            int value = l1.val + l2.val + isNeedPlusOne;            if (value >= 10) {                isNeedPlusOne = 1;                currentNode.next = new ListNode(value % 10);            } else {                isNeedPlusOne = 0;                currentNode.next = new ListNode(value);            }            currentNode = currentNode.next;            l1 = l1.next;            l2 = l2.next;        }        while (l1 != null) {            int value = l1.val + isNeedPlusOne;            if (value >= 10) {                isNeedPlusOne = 1;                currentNode.next = new ListNode(value % 10);            } else {                isNeedPlusOne = 0;                currentNode.next = new ListNode(value);            }            currentNode = currentNode.next;            l1 = l1.next;        }        while (l2 != null) {            int value = l2.val + isNeedPlusOne;            if (value >= 10) {                isNeedPlusOne = 1;                currentNode.next = new ListNode(value % 10);            } else {                isNeedPlusOne = 0;                currentNode.next = new ListNode(value);            }            currentNode = currentNode.next;            l2 = l2.next;        }        if (isNeedPlusOne == 1) {            currentNode.next = new ListNode(1);        }        return head.next;    }}

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