[leetcode]2. Add Two Numbers两数相加

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

　　

题目：

用两个链表表示的两个数，求相加之和

Solution1:  For given two lists, keep looping where either one isn't null. When ListNode comes to null, consider its val as 0.  

                For result list, use a dummy node.

     

   

code:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 /*
10  Time Complexity:  O(max(l1,l2))
11  Space Complexity: O(max(l1,l2)) 
12 */
13 class Solution {
14     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
15              ListNode dummy = new ListNode(-1);
16              ListNode pre = dummy;
17              ListNode p1 = l1;
18              ListNode p2 = l2; 
19              int carry = 0;    
20              while(p1 != null || p2 != null){
21                 int x = (p1 != null) ? p1.val : 0; 
22                 int y = (p2 != null) ? p2.val : 0;
23                 int sum = carry + x + y; 
24                 carry = sum / 10;
25                 // generate the result list
26                 pre.next = new ListNode(sum % 10);    
27                  // move three lists pointers  
28                 pre = pre.next;
29                 if(p1 != null) p1 = p1.next;
30                 if(p2 != null) p2 = p2.next;
31              }
32              // in case last two digit sum > 10 
33              if(carry > 0){
34                  pre.next = new ListNode(carry);
35              }
36              return dummy.next;     
37     }
38 }