Leetcode-2 ADD two Numbers calculate the list of two pairs of problems



1. Description of the problem:

You are given, linked, lists representing, and non-negativenumbers. The digits is stored in reverse order and each of the their nodes containa a single digit. ADD the numbers and return it as a linked list.

Input: (2, 4, 3) + (5, 6, 4)
Output:7, 0, 8

Definition of a linked list:

/** * Definition for singly-linked list. * public class ListNode {*     int val; *     ListNode Next; *     listnode (int x) {*         val = x; *         next = nul L *     } * } */

Here are a few things to consider: Considering that two linked lists are inconsistent in length, and that the list is still rounded up.

Solution One: The more tedious classification discussion method, avoids the use

public class Solution {public ListNode addtwonumbers (listnode L1, ListNode L2) {if ((l1==null) | |        (L2==null))        {return null;        } int temp_val = L1.val + l2.val;        int go = TEMP_VAL/10;         ListNode result = new ListNode (TEMP_VAL%10);        L1 = L1.next;        L2 = L2.next;        ListNode temp = result;            while ((L1!=null) && (l2!=null)) {temp_val = L1.val + L2.val + go;            ListNode temp2 = new ListNode (TEMP_VAL%10);             Temp.next = Temp2;            temp = TEMP2;            L1 = L1.next;            L2 = L2.next;        Go = TEMP_VAL/10;            } while (l1!=null) {temp_val = L1.val + go;            ListNode temp2 = new ListNode (TEMP_VAL%10);             Temp.next = Temp2;            temp = TEMP2;            L1 = L1.next;        Go = TEMP_VAL/10;            } while (l2!=null) {temp_val = L2.val + go; LiStnode temp2 = new ListNode (TEMP_VAL%10);             Temp.next = Temp2;            temp = TEMP2;            L2 = L2.next;        Go = TEMP_VAL/10;        } if (Go! = 0) {Temp.next = new ListNode (go);    } return result; }}

Solution Two: The better solution, but the running time on the Leetcode is inferior to the former

public class Solution {public ListNode addtwonumbers (listnode L1, ListNode L2) {    int carry = 0;//denotes carry    listnode hea D = new ListNode (0);//First node, refer to C + + post-tail node    listnode temp = head;    Loop until the two-linked list is traversed completely before exiting while    ((L1! = null) | | (L2! = null))    {    //Only a few factor sizes to be determined;    int first = (L1! = null)? l1.val:0;//use the IF statement also    int second = (L2! = null)? l2.val:0;
   int result = first + second + carry;//each cycle evaluates    carry = RESULT/10;    ListNode pnode = new ListNode (result%);    Temp.next = Pnode;    temp = pnode;//Prepares the next loop        if (L1! = null) {L1 = L1.next;}        if (L2! = null) {L2 = L2.next;}        }    Also consider the case where CArray is not equal to 0 and still has rounding    if (Carry > 0)    {    temp.next = new ListNode (carry);    }        Return head.next;//Note This return value    }}

Leetcode-2 ADD two Numbers calculate the list of two pairs of problems