从给定的字符串找到方向

给定仅包含L和R的字符串，分别代表左旋转和右旋转。任务是找到枢轴的最终方向(即N / E / S / W)。让枢轴在指南针中指向北(N)。

例子：

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

方法：

1. 使用一个计数器，该计数器在看到R时递增，而在看到L时递减。
2. 最后，在计数器上使用模数获取方向。
3. 如果计数为负，则方向将不同。还要检查代码是否为负。

下面是上述方法的实现：

C++

// CPP implementation of above approach
#include 
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 == 10):
            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

 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

    // 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.

输出

W
S

时间复杂度： O(N)，其中N是字符串的长度
辅助空间： O(1)

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