两个链表相加的和445. Add Two Numbers II

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题目：

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    /*    解题思路：将链表存入集合中，然后相加即可，定义一个变量flage记录进位    */    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode result=new ListNode(0);//头结点        LinkedList<Integer> list1=new LinkedList<Integer>();        LinkedList<Integer> list2=new LinkedList<Integer>();        while(l1!=null){            list1.add(l1.val);            l1=l1.next;        }//while        while(l2!=null){            list2.add(l2.val);            l2=l2.next;        }//while        int flage=0;        //头插法建立链表,集合是先进后出的方式        while(list1.size()!=0&&list2.size()!=0){            int num=list1.removeLast()+list2.removeLast()+flage;            flage=num/10;            num=num%10;            ListNode temp=new ListNode(num);            temp.next=result.next;            result.next=temp;        }//while        while(list1.size()!=0){            int num=list1.removeLast()+flage;            flage=num/10;            num=num%10;            ListNode temp=new ListNode(num);            temp.next=result.next;            result.next=temp;        }        while(list2.size()!=0){            int num=list2.removeLast()+flage;            flage=num/10;            num=num%10;            ListNode temp=new ListNode(num);            temp.next=result.next;            result.next=temp;        }        //考虑进位的时候，两个集合均为空的情况，例如{5}  {5}相加之后需要进位1        if(flage!=0){            ListNode temp=new ListNode(flage);            temp.next=result.next;            result.next=temp;        }        return result.next;    }}

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