Add Two Numbers

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

原本以为很简单的一道题目，竟然也折腾了一个多小时。

折腾的原因在于对Java对象引用的不熟悉。

<pre name="code" class="java">ListNode L1=new ListNode(1);ListNode P1=L1;P1.next=new ListNode(l1.next.val+l2.next.val);P1=P1.next;

这样做才能正常地把L1给链接起来，而我一开始犯了个非常低级的错误就是对P1进行了重新赋值。

ListNode L1=new ListNode(1);ListNode P1=L1;P1=new ListNode(l1.next.val+l2.next.val);P1=P1.next;

这样犯错的原因是把C++的指针给误理解过来了，理解成P1是指针，所以P1改变了后L1也跟着改变。

其实对P1进行重新赋值后，P1就指向了另外一个对象，跟L1就没关系了。

其他就是问题中的其他没考虑到的条件。

上代码

class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}}public class Solution {public static void main(String[] args) {ListNode l1 = new ListNode(0), l2 = new ListNode(0);ListNode P1 = l1, P2 = l2;// l1,P2=l2;int[] l1val = { 1, 2 };// , 9, 9, 9, 9, 8, 9, 9, 9 };int[] l2val = { 3, 5 };// , 2 };for (int indexL1val = 0; indexL1val < l1val.length; ++indexL1val) {P1.val = l1val[indexL1val];P1.next = new ListNode(0);P1 = P1.next;}for (int indexL2val = 0; indexL2val < l2val.length; ++indexL2val) {P2.val = l2val[indexL2val];P2.next = new ListNode(0);P2 = P2.next;}ListNode resultListNode = new Solution().addTwoNumbers(l1, l2);for (; resultListNode != null; resultListNode = resultListNode.next)System.out.println(resultListNode.val);}private ListNode addTwoNumbers(ListNode l1, ListNode l2) {int isBigTen = 0;ListNode L = new ListNode(0);ListNode P = L;ListNode l1tmp = l1, l2tmp = l2;for (; l1tmp != null && l2tmp != null; l1tmp = l1tmp.next, l2tmp = l2tmp.next) {P.val = (l1tmp.val + l2tmp.val + isBigTen);isBigTen = 0;// 记得消去if (P.val >= 10) {P.val = P.val - 10;isBigTen = 1;}if (l1tmp.next != null && l2tmp.next != null) {P.next = new ListNode(0);P = P.next;}}while(l1tmp!=null){P.next=new ListNode(l1tmp.val+isBigTen);isBigTen=0;P=P.next;if(P.val>=10){    P.val-=10;    isBigTen=1;}l1tmp=l1tmp.next;}while(l2tmp!=null){P.next=new ListNode(l2tmp.val+isBigTen);isBigTen=0;P=P.next;if(P.val>=10){    P.val-=10;    isBigTen=1;}l2tmp=l2tmp.next;}if(isBigTen==1){P.next=new ListNode(1);}return L;}}