2. Add Two Numbers

Problem:

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.

7/22/2018 update:

It it different from merge two sorted list. This one uses || the other one uses &&.

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
  ListNode head = dummy;
  int carry = 0;

  while (l1 != null || l2 != null || carry == 1) {
   int num1 = l1 == null ? 0 : l1.val;
   int num2 = l2 == null ? 0 : l2.val;
   int num = num1 + num2 + carry;
   carry = num / 10;
   head.next = new ListNode(num%10);
   head = head.next;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
  }
  return dummy.next;
    }
}

--------------------------------------------------------------------------------------------------------------------------
Analysis:
How to handle carry is a bit tricky. New carry is the sum/10. And the new value is sum%10.

Solution:

 /**  
  * 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 dummy = new ListNode(0);  
     ListNode sum = new ListNode(0);  
     dummy = sum;  
     int carry = 0;  
     while (l1 != null || l2 != null) {  
       int val1 = (l1 == null) ? 0 : l1.val;  
       int val2 = (l2 == null) ? 0 : l2.val;  
       int val = val1 + val2 + carry;  
       carry = val/10;  
       sum.next = new ListNode(val%10);  
       sum = sum.next;  
       if (l1 != null) l1 = l1.next;  
       if (l2 != null) l2 = l2.next;  
     }  
     if (carry != 0) {  
       sum.next = new ListNode(1);  
     }  
     return dummy.next;  
   }  
 }