LeetCode 2 Add Two Numbers

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

     解法思路：这道题不难，基本是链表操作。主要是要考虑较多的边界情况。在这里，我设置两个标志位end1、end2用于记录链是否到末尾。

边界情况如下：

(1).第一链小于第二链的情况

第一链到末尾，按照第二链的值继续做加运算。

(2).第一链大于第二链的情况

第二链到末尾，按照第一链的值继续做加运算。

(3)每次链表遍历的时候，都要考虑是否到末尾。

(4)进位情况应该记录在下一个链next中，无需再设标志位。

     代码如下：

1.链表结构

class ListNode {    int val;   ListNode next;    ListNode(int x) { val = x; } }

2.算法

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode l3 = new ListNode(0);ListNode t1 = l1;ListNode t2 = l2;ListNode t3 =l3;int end1 = 0;//结束位 1表示结束  0表示未结束int end2 = 0;while(t1!=null || t2!=null){if(end1 == 0 && end2 == 0){l3.val += t1.val+t2.val;if(l3.val>=10){l3.val = l3.val-10;l3.next = new ListNode(1); t1 = t1.next; t2 = t2.next; l3 = l3.next; if(end1!=1){ if(t1 == null){end1 = 1; } }if(end2!=1){ if(t2 == null){ end2 = 1; } } if(end1 == 1 && end2 ==1){ break; } continue;}}if(end1 == 0 && end2 == 1){l3.val += t1.val;if(l3.val>=10){l3.val = l3.val-10;l3.next = new ListNode(1);if(end1!=1){ t1 = t1.next;}    if(end2!=1){    t2 = t2.next;    } l3 = l3.next; if(end1!=1){ if(t1 == null){end1 = 1; } }if(end2!=1){ if(t2 == null){ end2 = 1; } } if(end1 == 1 && end2 ==1){ break; } continue;}}if (end1 == 1 && end2 == 0) {l3.val += t2.val;if(l3.val>=10){l3.val = l3.val-10;l3.next = new ListNode(1);if(end1!=1){ t1 = t1.next;}    if(end2!=1){    t2 = t2.next;    } l3 = l3.next; if(end1!=1){ if(t1 == null){end1 = 1; } }if(end2!=1){ if(t2 == null){ end2 = 1; } } if(end1 == 1 && end2 ==1){ break; } continue;}}if(end1!=1){ t1 = t1.next;}    if(end2!=1){    t2 = t2.next;    } if(end1!=1){ if(t1 == null){end1 = 1; } } if(end2!=1){ if(t2 == null){ end2 = 1; } } if(end1 == 1 && end2 == 1){ break; } l3.next = new ListNode(0);l3 = l3.next;}return t3;    }