[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;     
    }
}