加法合并链表

题目：

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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

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

思路：

没有要求时间空间复杂度，单纯链表操作。而且给出了Node的定义。建一个临时的head，指向第一个Node，根据加法算出后面的Node。再返回head.next即可。注意一些特殊情况以免漏掉一些case。

代码：（有点冗余，但是勉强ac了）

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
   public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
       ListNode head = new ListNode(0);
       ListNode temp = head;
       int flag = 0;
       while(l1!=null&&l2!=null){
           if((l1.val+l2.val+flag)>9){
               temp.next = new ListNode(l1.val+l2.val-10+flag);
               temp = temp.next;
               flag = 1;
           }else {
               temp.next = new ListNode(l1.val+l2.val+flag);
               temp = temp.next;
               flag = 0;
           }
           l1 = l1.next;
           l2 = l2.next;
       }
       while(l1!=null){
           if((l1.val+flag)>9){
                temp.next = new ListNode(l1.val+flag-10);
                temp = temp.next;
                flag = 1;
           }
           else{
               temp.next = new ListNode(l1.val+flag);
                temp = temp.next;
                flag = 0;
           }
           l1 = l1.next;
          
       } 
        while(l2 !=null){
           if((l2.val+flag)>9){
                temp.next = new ListNode(l2.val+flag-10);
                temp = temp.next;
                flag = 1;
           }
           else{
               temp.next = new ListNode(l2.val+flag);
                temp = temp.next;
                flag = 0;
           }
           l2 = l2.next;
           
       }
       if(flag == 1){
           temp.next = new ListNode(1);
       }
       return head.next;
   }
}