从给定的字符串中找到方向
给定一个只包含左旋转和右旋转的字符串。任务是找到枢轴的最终方向(即北/东/南/西)。让一个支点指向罗盘上的北。
示例:
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
进场:
- 使用一个在看到 R 时递增，在看到 l 时递减的计数器
- 最后，在计数器上使用模来获得方向。
- 如果计数为负，则方向将不同。检查代码是否也是负数。
下面是上述方法的实现:
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 语言(一种计算机语言，尤用于创建网站)
// 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));
}
}
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?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
?>
java 描述语言
<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>
Output
W
S
时间复杂度: O(N)其中 N 为弦的长度 T3】辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
