LeetCode【2】. Add Two Numbers--java实现

Add Two Numbers

        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

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(in x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode slist = new ListNode(0);
        ListNode clist = slist;
        ListNode nlist = new ListNode(0);
        //int sval = 0;
        int flag = 0; // 进位
        
        //1. if First Node of l1 or l2 is null 
        if(l1==null||l2==null){
           return (l1==null)?((l2==null)?(slist):(l2)):(l1);
        }
        //2.1 当l1，l2都非链尾时，loop
        while(true)
        {
            clist.val = (l1.val + l2.val + flag)%(10);
            flag = (l1.val + l2.val + flag)/10;
            //next node
            l1 = l1.next;
            l2  = l2.next;
            
            //2.1.1  若任意一个为链尾，则跳出
            if(l1==null||l2==null){
                 break;
            }else{
            clist.next= new ListNode(0);
            clist =clist.next;
            }
        };//while
        
        //2.2 如果两个同时为链尾时
        if(l1==null&&l2==null)
        {
            //2.2.1 若两个为链尾且有进位，结果需进位
            if(flag==1){
               nlist = new ListNode(flag);
               clist.next = nlist;
            }else{
                return slist;
            }
        }else //2.2 一个到达链尾、一个还未 
        {
            ListNode onelist = new ListNode(0);
            if(l1==null)
            {onelist = l2;
            }else
            {onelist = l1; }
            while(onelist!= null)
            {
              
              clist.next =  new ListNode(0);
              clist = clist.next;
              clist.val = (onelist.val + flag)%10;
              flag = (onelist.val + flag)/10;
              onelist = onelist.next;
            }
            //2.2.1 当另外一个也到达链尾，判断是否有进位
            if(flag==1)
            {
                clist.next = new ListNode(flag);
            }
        }
        
        return slist;
        
    }
}