You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

Solution1：不用分配多余的空间，但是会改变原先lists中的值。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        if(l1==null&&l2==null) return null;
        if(l1==null) return l2;
        if(l2==null) return l1;
        ListNode ptr1 = l1, ptr2 = l2;
        int c = 0; 
        ListNode prev=l1;//reserve the previous value in the generated list
        while(ptr1!=null && ptr2!=null){
            ptr1.val = ptr1.val+ptr2.val+c;
            c = ptr1.val/10;
            ptr1.val = ptr1.val%10;
            prev = ptr1;
            ptr1=ptr1.next;
            ptr2=ptr2.next;
        }
        if(ptr2!=null) {    //ptr1==null
            prev.next = ptr2;
            ptr1 = ptr2;
            ptr2=null;  //must do this for maintaining end condition
            
        }
        while(c!=0&&ptr1!=null){
            ptr1.val =ptr1.val+c;
            c = ptr1.val/10;
            ptr1.val = ptr1.val%10;
            prev = ptr1;
            ptr1 = ptr1.next;
        }
       
        if(c!=0&&ptr1==null&&ptr2==null) {  
            ListNode newNode =new ListNode(1);
            prev.next = newNode;
            ptr1 = newNode;
        }
        return l1;
    }
}

Solution 2: 不会改变原lists中的值，但需建立一个新list

 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        if(l1==null&&l2==null) return null;
        if(l1==null) return l2;
        if(l2==null) return l1;
        ListNode head = new ListNode(0);
        int c = 0; 
        ListNode prev=head;//reserve the previous value in the generated list
        while(l1!=null || l2!=null){
            ListNode cur = new ListNode(0);
            if(l1!=null){
                cur.val += l1.val;
                l1=l1.next;
            }
             if(l2!=null){
                cur.val += l2.val;
                l2=l2.next;
            }
            cur.val += c;
            c = cur.val/10;
            cur.val = cur.val%10;
            prev.next = cur;
            prev = cur;
        }
       
        if(c!=0) {  
            ListNode newNode =new ListNode(1);
            prev.next = newNode;
        }
        return head.next;
    }
}

第13题 Add Two Numbers