| 标题： | Add Two Numbers |
| 通过率： | 22.6% |
| 难度： | 中等 |
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
本题是比较弱的。要注意的就是两个链表的长度不一样，在这里我还考虑了余数大于十的问题。
代码如下：
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 14 int list1num=-1,list2num=-1; 15 ListNode res=new ListNode(1); 16 ListNode result=res; 17 int carry=0,sum=0; 18 boolean flag=true; 19 while(flag){ 20 if(l1!=null&l2!=null){ 21 list1num=l1.val+carry; 22 list2num=l2.val; 23 carry=0; 24 carry=(list1num+list2num)/10; 25 sum=(list1num+list2num)%10; 26 ListNode temp=new ListNode(sum); 27 res.next=temp; 28 res=temp; 29 l1=l1.next; 30 l2=l2.next; 31 } 32 else if(l1!=null&&l2==null){ 33 list1num=l1.val+carry; 34 carry=0; 35 carry=list1num/10; 36 sum=list1num%10; 37 ListNode temp=new ListNode(sum); 38 res.next=temp; 39 res=temp; 40 l1=l1.next; 41 } 42 else if(l1==null&&l2!=null){ 43 list2num=l2.val+carry; 44 carry=0; 45 carry=list2num/10; 46 sum=list2num%10; 47 ListNode temp=new ListNode(sum); 48 res.next=temp; 49 res=temp; 50 l2=l2.next; 51 } 52 else { 53 flag=false; 54 55 } 56 57 } 58 while((carry/10!=0)||(carry%10!=0)){ 59 ListNode temp=new ListNode(carry%10); 60 carry=carry/10; 61 res.next=temp; 62 res=temp; 63 } 64 return result.next; 65 66 } 67 }