[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:

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) { val = x; }
  * }
  */
 /*
  Time Complexity:  O(max(l1,l2))
  Space Complexity: O(max(l1,l2))
 */
 class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
              ListNode dummy = new ListNode(-1);
              ListNode pre = dummy;
              ListNode p1 = l1;
              ListNode p2 = l2;
              int carry = 0;
              while(p1 != null || p2 != null){
                 int x = (p1 != null) ? p1.val : 0;
                 int y = (p2 != null) ? p2.val : 0;
                 int sum = carry + x + y;
                 carry = sum / 10;
                 // generate the result list
                 pre.next = new ListNode(sum % 10);
                  // move three lists pointers
                 pre = pre.next;
                 if(p1 != null) p1 = p1.next;
                 if(p2 != null) p2 = p2.next;
              }
              // in case last two digit sum > 10
              if(carry > 0){
                  pre.next = new ListNode(carry);
              }
              return dummy.next;
     }
 }