大佬解法

/**
 * 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) {
        ListNode pre = new ListNode(0);
        ListNode cur = pre;
        int carry = 0;
        while(l1 != null || l2 != null) {
            int x = l1 == null ? 0 : l1.val;
            int y = l2 == null ? 0 : l2.val;
            int sum = x + y + carry;
            
            carry = sum / 10;
            sum = sum % 10;
            cur.next = new ListNode(sum);

            cur = cur.next;
            if(l1 != null)
                l1 = l1.next;
            if(l2 != null)
                l2 = l2.next;
        }
        if(carry == 1) {
            cur.next = new ListNode(carry);
        }
        return pre.next;
    }
}

自己的解法

写的跟屎一样,虽然击败100%Java用户,但是写的实在是太过于复杂

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int sum=0;
        ListNode ans=new ListNode();
        ListNode p=ans,ptr=null;
        for(;;){
            if(l1!=null&&l2!=null)
            {
                sum=l1.val+l2.val;
                if(p.val==1){
                    sum+=1;
                }
                if(sum>=10){
                    p.next=new ListNode(1);
                    p.val=sum%10;
                }else{
                    p.val=sum;
                    if(l1.next!=null||l2.next!=null)//防止产生前导0
                    {
                        p.next=new ListNode();
                    }
                }
                l1=l1.next;
                l2=l2.next;
                ptr=l1==null?l2:l1;
                p=p.next;
            }
            else if(ptr!=null){
            
                int null_sum=0;
                if(p.val==1){
                    null_sum=ptr.val+p.val;
                    if(null_sum>=10){
                        p.next=new ListNode(1);
                        p.val=null_sum%10;

                    }else{
                        p.val=null_sum;
                        if(ptr.next!=null){//防止产生前导0
                            p.next=new ListNode();
                        }
                        
                    }
                }
                else{
                    p.val=ptr.val;
                    if(ptr.next!=null){//防止产生前导0
                        p.next=new ListNode();
                    }
                }
                //往后遍历
                ptr=ptr.next;
                p=p.next;
            }else{
                break;
            }
        }
        return ans; 
    }

}