NO.2 Add Two Numbers

使用递归算法

/**
 * 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) {
        return add(l1, l2, 0);
    }
   public ListNode add(ListNode l1,ListNode l2,int carry)
     {
          if(l1==null && l2==null){
                return carry == 0? null : new ListNode(carry);
            }
            if(l1==null && l2!=null){
                l1 = new ListNode(0);
            }
            if(l2==null && l1!=null){
                l2 = new ListNode(0);
            }
            int sum = l1.val + l2.val + carry;
            ListNode curr = new ListNode(sum % 10);
            curr.next = add(l1.next, l2.next, sum / 10);
            return curr;
     }
}

偷懒相加法，超过long的范围会失败

/**
 * 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) {
        if(l1==null&&l2==null){
             return null;
         }
         //链表转long型
         long num1 = listToLong(l1);
         long num2 = listToLong(l2);
         return longToList(num1+num2);
    }
   //链表转数据
    public Long listToLong(ListNode l)
    {
        long num=0;
        long curr=1;
        int i=0;
        while(l!=null)
        {
            num=num+l.val*curr;
            curr=curr*10;
            l=l.next;
        }
        return num;
    }
    
    //数据转链表
    public ListNode longToList(Long num){       
        ListNode l3 = new ListNode(-1);
        l3.next = null;
        ListNode c = l3;
        c.val=(int)(num%10);
        num = num/10;
        while(num>0){       
            ListNode cnext = new ListNode((int)(num%10));
            cnext.next=null;
            c.next=cnext;
            num = num/10;
            c=c.next;
        }
        return l3;
    }
}