Leetcode-add Numbers

ADD Numbers

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

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

Two numbers exist in two single-linked lists, asking for their and, as a result, there is a linked list.

Java programs

/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution { PublicListNode addtwonumbers (listnode L1, ListNode L2) {ListNode Headnode=NewListNode (0); ListNode CurrentNode=Headnode; intCArray = 0;  while(L1! =NULL&& L2! =NULL) {CArray+=L1.val; L1=L1.next; CArray+=L2.val; L2=L2.next; Currentnode.next=NewListNode (carray%10); CurrentNode=Currentnode.next; CArray/=10; }        if(l1!=NULL){                            while(l1!=NULL) {CArray+=L1.val; L1=L1.next; Currentnode.next=NewListNode (carray%10); CurrentNode=Currentnode.next; CArray/=10; }            }                    if(l2!=NULL){                 while(l2!=NULL) {CArray+=L2.val; L2=L2.next; Currentnode.next=NewListNode (carray%10); CurrentNode=Currentnode.next; CArray/=10; }            }        if(carray==1) {Currentnode.next=NewListNode (1); CurrentNode=Currentnode.next; }       returnHeadnode.next; }}

First two nodes, one pointing to the head node, one to the current running node

Simultaneously traverse two single linked list, corresponding position sum, carray%10 as new node value, CARRAY/10 is to carry

When a linked list is empty, the description is two different lengths of the number added, the current node to consider rounding the case, the non-empty nodes are linked together.

The upper while is the operation when the two linked list is not empty, and then joins the non-empty list

You can combine the two together.

/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution { PublicListNode addtwonumbers (listnode L1, ListNode L2) {ListNode Headnode=NewListNode (0); ListNode CurrentNode=Headnode; intCArray = 0;  while(L1! =NULL|| L2! =NULL){              if(l1!=NULL) {CArray+=L1.val; L1=L1.next; }            if(l2!=NULL) {CArray+=L2.val; L2=L2.next; } Currentnode.next=NewListNode (carray%10); CurrentNode=Currentnode.next; CArray/=10; }                if(carray==1) {Currentnode.next=NewListNode (1); CurrentNode=Currentnode.next; }       returnHeadnode.next; }}

This is a much simpler procedure.

Python program

#Definition for singly-linked list.#class ListNode (object):#def __init__ (self, x):#self.val = x#Self.next = Noneclasssolution (object):defaddtwonumbers (self, L1, L2):""": Type L1:ListNode:type l2:ListNode:rtype:ListNode"""Carry=0 Head=listnode (0) CurrentNode=Head whileL1orL2:ifL1:carry+=l1.val L1=L1.nextifL2:carry+=L2.val L2=L2.next Currentnode.next= ListNode (carry%10) CurrentNode=Currentnode.next Carry= CARRY//10ifCarry==1: Currentnode.next= ListNode (1) CurrentNode=Currentnode.nextreturnHead.next

Leetcode-add Numbers