Encuentra la dirección de la string dada

Posted on by Rudeus Greyrat

Dada una string que contiene solo L y R, que representa la rotación a la izquierda y la rotación a la derecha, respectivamente. La tarea es encontrar la dirección final del pivote (es decir, N/E/S/W). Deje que un pivote apunte hacia el norte (N) en una brújula. 

Ejemplos: 

Input: str = "LLRLRRL"
Output: W
In this input string we rotate pivot to left
when a L char is encountered and right when 
R is encountered. 

Input: str = "LL"
Output: S

Acercarse:  

  1. Use un contador que incremente al ver R y disminuya al ver L.
  2. Finalmente, use módulo en el mostrador para obtener la dirección.
  3. Si el conteo es negativo, las direcciones serán diferentes. Verifique el código para ver si es negativo también.

A continuación se muestra la implementación del enfoque anterior: 

C++

// CPP implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the final direction
string findDirection(string s)
{
    int count = 0;
    string d = "";
 
    for (int i = 0; i < s.length(); i++) {
 
        if (s[0] == '\n')
            return NULL;
 
        if (s[i] == 'L')
            count--;
        else {
            if (s[i] == 'R')
                count++;
        }
    }
 
    // if count is positive that implies
    // resultant is clockwise direction
    if (count > 0) {
 
        if (count % 4 == 0)
            d = "N";
        else if (count % 4 == 1)
            d = "E";
        else if (count % 4 == 2)
            d = "S";
        else if (count % 4 == 3)
            d = "W";
    }
 
    // if count is negative that implies
    // resultant is anti-clockwise direction
    if (count < 0) {
 
        if (count % 4 == 0)
            d = "N";
        else if (count % 4 == -1)
            d = "W";
        else if (count % 4 == -2)
            d = "S";
        else if (count % 4 == -3)
            d = "E";
    }
    return d;
}
 
// Driver code
int main()
{
    string s = "LLRLRRL";
    cout << (findDirection(s)) << endl;
 
    s = "LL";
    cout << (findDirection(s)) << endl;
}
 
// This code is contributed by
// SURENDRA_GANGWAR

Java

// Java implementation of above approach
import java.util.*;
 
class GFG {
 
    // Function to find the final direction
    static String findDirection(String s)
    {
        int count = 0;
        String d = "";
 
        for (int i = 0; i < s.length(); i++) {
 
            if (s.charAt(0) == '\n')
                return null;
 
            if (s.charAt(i) == 'L')
                count--;
            else {
                if (s.charAt(i) == 'R')
                    count++;
            }
        }
 
        // if count is positive that implies
        // resultant is clockwise direction
        if (count > 0) {
 
            if (count % 4 == 0)
                d = "N";
            else if (count % 4 == 1)
                d = "E";
            else if (count % 4 == 2)
                d = "S";
            else if (count % 4 == 3)
                d = "W";
        }
 
        // if count is negative that implies
        // resultant is anti-clockwise direction
        if (count < 0) {
 
            if (count % 4 == 0)
                d = "N";
            else if (count % 4 == -1)
                d = "W";
            else if (count % 4 == -2)
                d = "S";
            else if (count % 4 == -3)
                d = "E";
        }
        return d;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "LLRLRRL";
        System.out.println(findDirection(s));
 
        s = "LL";
        System.out.println(findDirection(s));
    }
}

Python3

# Python3 implementation of
# the above approach
 
# Function to find the
# final direction
 
 
def findDirection(s):
 
    count = 0
    d = ""
 
    for i in range(len(s)):
        if (s[i] == 'L'):
            count -= 1
        else:
            if (s[i] == 'R'):
                count += 1
 
    # if count is positive that
    # implies resultant is clockwise
    # direction
    if (count > 0):
        if (count % 4 == 0):
            d = "N"
        elif (count % 4 == 1):
            d = "E"
        elif (count % 4 == 2):
            d = "S"
        elif (count % 4 == 3):
            d = "W"
 
    # if count is negative that
    # implies resultant is anti-
    # clockwise direction
    if (count < 0):
        count *= -1
        if (count % 4 == 0):
            d = "N"
        elif (count % 4 == 1):
            d = "W"
        elif (count % 4 == 2):
            d = "S"
        elif (count % 4 == 3):
            d = "E"
 
    return d
 
 
# Driver code
if __name__ == '__main__':
 
    s = "LLRLRRL"
    print(findDirection(s))
 
    s = "LL"
    print(findDirection(s))
 
# This code is contributed by 29AjayKumar

C#

// C# implementation of above approach
using System;
 
class GFG {
 
    // Function to find the final direction
    static String findDirection(String s)
    {
        int count = 0;
        String d = "";
 
        for (int i = 0; i < s.Length; i++) {
 
            if (s[0] == '\n')
                return null;
 
            if (s[i] == 'L')
                count--;
            else {
                if (s[i] == 'R')
                    count++;
            }
        }
 
        // if count is positive that implies
        // resultant is clockwise direction
        if (count > 0) {
 
            if (count % 4 == 0)
                d = "N";
            else if (count % 4 == 1)
                d = "E";
            else if (count % 4 == 2)
                d = "S";
            else if (count % 4 == 3)
                d = "W";
        }
 
        // if count is negative that implies
        // resultant is anti-clockwise direction
        if (count < 0) {
 
            if (count % 4 == 0)
                d = "N";
            else if (count % 4 == -1)
                d = "W";
            else if (count % 4 == -2)
                d = "S";
            else if (count % 4 == -3)
                d = "E";
        }
        return d;
    }
 
    // Driver code
    public static void Main()
    {
        String s = "LLRLRRL";
        Console.WriteLine(findDirection(s));
 
        s = "LL";
        Console.WriteLine(findDirection(s));
    }
}
 
// This code is contributed by Shashank

PHP

<?php
// PHP implementation of above approach
 
// Function to find the final direction
function findDirection($s)
{
    $count = 0;
    $d = "";
 
    for ($i = 0;
         $i < strlen($s); $i++)
    {
        if ($s[0] == '\n')
            return null;
 
        if ($s[$i] == 'L')
            $count -= 1;
        else
        {
            if ($s[$i] == 'R')
                $count += 1;
        }
    }
             
    // if count is positive that implies
    // resultant is clockwise direction
    if ($count > 0)
    {
        if ($count % 4 == 0)
            $d = "N";
        else if ($count % 4 == 1)
            $d = "E";
        else if ($count % 4 == 2)
            $d = "S";
        else if ($count % 4 == 3)
            $d = "W";
    }
     
    // if count is negative that
    // implies resultant is
    // anti-clockwise direction
    if ($count < 0)
    {
        if ($count % 4 == 0)
            $d = "N";
        else if ($count % 4 == -1)
            $d = "W";
        else if ($count % 4 == -2)
            $d = "S";
        else if ($count % 4 == -3)
            $d = "E";
    }
    return $d;
}
 
// Driver code
$s = "LLRLRRL";
echo findDirection($s)."\n";
 
$s = "LL";
echo findDirection($s)."\n";
 
// This code is contributed
// by ChitraNayal   
?>

Javascript

<script>
    // Javascript implementation of above approach
     
    // Function to find the final direction
    function findDirection(s)
    {
        let count = 0;
        let d = "";
  
        for (let i = 0; i < s.length; i++) {
  
            if (s[0] == '\n')
                return null;
  
            if (s[i] == 'L')
                count--;
            else {
                if (s[i] == 'R')
                    count++;
            }
        }
  
        // if count is positive that implies
        // resultant is clockwise direction
        if (count > 0) {
  
            if (count % 4 == 0)
                d = "N";
            else if (count % 4 == 1)
                d = "E";
            else if (count % 4 == 2)
                d = "S";
            else if (count % 4 == 3)
                d = "W";
        }
  
        // if count is negative that implies
        // resultant is anti-clockwise direction
        if (count < 0) {
  
            if (count % 4 == 0)
                d = "N";
            else if (count % 4 == -1)
                d = "W";
            else if (count % 4 == -2)
                d = "S";
            else if (count % 4 == -3)
                d = "E";
        }
        return d;
    }
     
    let s = "LLRLRRL";
    document.write(findDirection(s) + "</br>");
 
    s = "LL";
    document.write(findDirection(s));
         
        // This code is contributed by divyeshrabadiya07.
</script>
Producción

W
S

Complejidad de tiempo: O(N) donde N es la longitud de la string
Espacio auxiliar: O(1)

Publicación traducida automáticamente

Artículo escrito por kalyan y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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