167. Add Two Numbers【LintCode by java】

Description

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

Example

Given 7->1->6 + 5->9->2. That is, 617 + 295.

Return 2->1->9. That is 912.

Given 3->1->5 and 5->9->2, return 8->0->8

解题：题目的意思是，给两个链表，倒序表示一个若干位的数。要求返回这两个数的和，并且格式是题中规定的链表，也是倒序。思路很清晰，从头到尾，一位一位地相加，并且用一个数来保存进位。每次相加的时候，都得考虑进位，把进位算在其中。当然，思路越清晰，代码可能看起来就比较笨重。代码如下：

 /**
  * Definition for ListNode
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */

 public class Solution {
     /**
      * @param l1: the first list
      * @param l2: the second list
      * @return: the sum list of l1 and l2
      */
     public ListNode addLists(ListNode l1, ListNode l2) {
         // write your code here
         ListNode head = new ListNode(0);
         ListNode p=head;
         ListNode p1 = l1;
         ListNode p2 = l2;
         int sum = 0;//保存每一位的和
         int more=0;//保存进位
         while(p1 != null && p2 != null){
             sum = p1.val + p2.val+more;
             more = sum / 10;
             sum = sum % 10;
             ListNode temp = new ListNode(sum);
             p.next = temp;
             p=p.next;
             p1 = p1.next;
             p2 = p2.next;
         }
         //如果more不为0
         sum = 0;
         while(p1 != null){
             sum = more + p1.val;
             more = sum / 10;
             sum = sum % 10;
             ListNode temp = new ListNode(sum);
             p.next = temp;
             p = p.next;
             p1 = p1.next;
         }
         while(p2 != null){
             sum = more + p2.val;
             more = sum / 10;
             sum = sum % 10;
             ListNode temp = new ListNode(sum);
             p.next = temp;
             p = p.next;
             p2 = p2.next;
         }
         if(more != 0){                //如果more不为0，那么最高位就是前面的进位
             ListNode temp = new ListNode(more);
             p.next = temp;
         }
         return head.next;
     }
 }

代码可能有点冗余，曾尝试改进，但发现还是这样写看起来比较清晰。如有错误，欢迎批评指正。