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I am a backend developer (PHP, Javascript, Laravel, Node ) , I know frontend (css, html, react, react-native) but it is not my passion, I have experience as a linux system administrator. I have deployed products in googlecloud, I have experience with integration and maintenance to mobile applications, in addition to working with database sql (mysql, postgresql) and nosql (mongodb)
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Programming Languages war
Programming Languages war
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Hey my friends, a new post about this war. I don't know which is better, but I'll try to figure it out. I hope y'all enjoy this series. Here we go!
This is a challenge of leetcode.com I made this challenge with python3, and next I made PHP and Javacript the same way.
Problem:
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807.
I receive two objects and I have to add, I used "while" to iterate the object and pass to list, it's easier manipulate the number of this way, I reverse the list with the method "reverse()", and turn list to string and then to integer, I add both numbers and reverse again and return the list converted to ListNode Object.
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def addTwoNumbers(self, l1, l2):
al = list()
bl = list()
while l1 or l2:
(al.append(l1.val) if l1 else None)
(bl.append(l2.val) if l2 else None)
l1 = (l1.next if l1 else None)
l2 =(l2.next if l2 else None)
al.reverse()
bl.reverse()
str1 = ''.join(map(str, al))
str2 = ''.join(map(str, bl))
result = int(str1) + int(str2)
result = list(str(result))
result.reverse()
return self.createList(ListNode(0), result, 0)
def createList(self, node, list, pos):
if node.val == 0:
if len(list) > pos:
node.val = list[pos]
self.createList(node, list, pos+1)
else:
if len(list) > pos:
node.next = ListNode(list[pos])
if(len(list) > pos+1):
self.createList(node.next, list, pos+1)
else:
return node
return node
sol = Solution()
a = sol.createList(ListNode(0), [2,4,3], 0)
b = sol.createList(ListNode(0), [5,6,4], 0)
solve = sol.addTwoNumbers(a, b)
print(solve)
I receive two objects and I have to add, I used "while" to iterate the object and to turn to array, it's easier to manipulate the number of this way. I try to use reverse here, that exist a method for that, but when I try to send my solution is not successful in all cases. For example an array with length 31, because when I turn this array in a number return a number like 1E+30, then I can't solve that, but I resolve the challenge the other way. I validate what an array is longer, so, I add but to left to right and validate if the result is gran than 9 and save the number in a new array and return the array converted to ListNode Object.
<?php
class ListNode
{
public $val = 0;
public $next = null;
public function __construct($val = 0, $next = null)
{
$this->val = $val;
$this->next = $next;
}
}
class Solution
{
public function addTwoNumbers($l1, $l2)
{
$al = array();
$bl = array();
while ($l1) {
$al[] = $l1->val;
$l1 = $l1->next ? $l1->next : null;
}
while ($l2) {
$bl[] = $l2->val;
$l2 = $l2->next ? $l2->next : null;
}
if (count($al) > count($bl) || count($al) == count($bl)) {
$pending = 0;
for ($i = 0; $i < count($al); $i++) {
$a = $al[$i];
$b = !empty($bl[$i]) ? $bl[$i] : 0;
$r = $a + $b + $pending;
if ($r > 9) {
$result[] = (strval($r))[1];
$pending = (strval($r))[0];
} else {
$result[] = strval($r);
$pending = 0;
}
}
if($pending){
$result[] = $pending;
}
} else {
$pending = 0;
for ($i = 0; $i < count($bl); $i++) {
$a = $bl[$i];
$b = !empty($al[$i]) ? $al[$i] : 0;
$r = $a + $b + $pending;
if ($r > 9) {
$result[] = (strval($r))[1];
$pending = (strval($r))[0];
} else {
$result[] = strval($r);
$pending = 0;
}
}
if($pending){
$result[] = $pending;
}
}
return $this->createList(new ListNode(0), $result, 0);
}
public function createList($node, $arr, $pos)
{
if ($node->val == 0 && $pos == 0) {
if (count($arr) > $pos) {
$node->val = $arr[$pos];
$this->createList($node, $arr, $pos + 1);
}
} else {
if (count($arr) > $pos) {
$node->next = new ListNode($arr[$pos]);
if (count($arr) > $pos + 1) {
$this->createList($node->next, $arr, $pos + 1);
} else {
return $node;
}
}
}
return $node;
}
}
$sol = new Solution();
$a = $sol->createList(new ListNode(0), [2,4,9], 0);
$b = $sol->createList(new ListNode(0), [5,6,4,9], 0);
$solve = $sol->addTwoNumbers($a, $b);
var_dump($solve);
?>
I receive two objects and I have to add, I used "while" to iterate the object and to turn to array, it's easier to manipulate the number of this way, I try to use reverse here, that exist a method for that, but when to try to send my solution is not successful in all cases. For example an array with length 31, because when I turn this array in a number return a number like 1E+30 the same that PHP, then I can't solve that, but I resolve the challenge the other way. I validate what an array is longer, so, I add but to left to right and validate if the result is gran than 9 and save the number in a new array and return the array converted to ListNode Object.
class ListNode {
constructor(val = 0, next = null) {
this.val = val === undefined ? 0 : val;
this.next = next === undefined ? null : next;
}
}
class Solution {
addTwoNumbers(l1, l2) {
let al = [];
let bl = [];
while (l1) {
al.push(l1.val);
l1 = l1.next ? l1.next : null;
}
while (l2) {
bl.push(l2.val);
l2 = l2.next ? l2.next : null;
}
let pending = 0;
let result = [];
if (al.length > bl.length || al.length == bl.length) {
for (let i = 0; i < al.length; i++) {
let a = al[i];
let b = bl[i] ? bl[i] : 0;
let r = a + b + parseInt(pending);
if (r > 9) {
result.push(String(r)[1]);
pending = String(r)[0];
} else {
result.push(String(r));
pending = 0;
}
}
if (pending) {
result.push(pending);
}
}
if (al.length < bl.length) {
for (let i = 0; i < bl.length; i++) {
let a = bl[i];
let b = al[i] ? al[i] : 0;
let r = a + b + parseInt(pending);
if (r > 9) {
result.push(String(r)[1]);
pending = String(r)[0];
} else {
result.push(String(r));
pending = 0;
}
}
if (pending) {
result.push(pending);
}
}
return this.createList(new ListNode(0), result, 0);
}
createList(node, arr, pos) {
if (node.val == 0 && pos == 0) {
if (arr.length > pos) {
node.val = arr[pos];
this.createList(node, arr, pos + 1);
}
} else {
if (arr.length > pos) {
node.next = new ListNode(arr[pos]);
if (arr.length > pos + 1) {
this.createList(node.next, arr, pos + 1);
} else {
return node;
}
}
}
return node;
}
}
const sol = new Solution();
const a = sol.createList(new ListNode(0), [2, 4, 3], 0);
const b = sol.createList(new ListNode(0), [5, 6, 4], 0);
const solve = sol.addTwoNumbers(a, b);
console.log(solve);
Conclusion
Structure: Python: I don't have problem to manipulate long integer, PHP and Javascript I have to change my first solution, I don't change my python way to be like PHP and Javascript because was a little slower in Python. Lines: Python: 47, PHP: 107, Javascript: 87 Time to execution: Python: 72 ms, PHP: 28 ms, Javascript: 144 ms Memory Usage: Python: 14.5 MB, PHP: 15.8 MB, Javascript: 46.6 MB
Would you like that I write about more challenges?
I hope you enjoy my post and remember that I am just a Dev like you!