Dada una string s1 y una string s2, escribe un fragmento para decir si s2 es una rotación de s1.
(por ejemplo, dado s1 = ABCD y s2 = CDAB, devuelve verdadero, dado s1 = ABCD y s2 = ACBD, devuelve falso)
Algoritmo: areRotations(str1, str2)
C++
// C++ program to check if two given strings
// are rotations of each other
# include <bits/stdc++.h>
using namespace std;
/* Function checks if passed strings (str1
and str2) are rotations of each other */
bool areRotations(string str1, string str2)
{
/* Check if sizes of two strings are same */
if (str1.length() != str2.length())
return false;
string temp = str1 + str1;
return (temp.find(str2) != string::npos);
}
/* Driver program to test areRotations */
int main()
{
string str1 = "AACD", str2 = "ACDA";
if (areRotations(str1, str2))
printf("Strings are rotations of each other");
else
printf("Strings are not rotations of each other");
return 0;
}
C
// C program to check if two given strings are rotations of
// each other
# include <stdio.h>
# include <string.h>
# include <stdlib.h>
/* Function checks if passed strings (str1 and str2)
are rotations of each other */
int areRotations(char *str1, char *str2)
{
int size1 = strlen(str1);
int size2 = strlen(str2);
char *temp;
void *ptr;
/* Check if sizes of two strings are same */
if (size1 != size2)
return 0;
/* Create a temp string with value str1.str1 */
temp = (char *)malloc(sizeof(char)*(size1*2 + 1));
temp[0] = '';
strcat(temp, str1);
strcat(temp, str1);
/* Now check if str2 is a substring of temp */
ptr = strstr(temp, str2);
free(temp); // Free dynamically allocated memory
/* strstr returns NULL if the second string is NOT a
substring of first string */
if (ptr != NULL)
return 1;
else
return 0;
}
/* Driver program to test areRotations */
int main()
{
char *str1 = "AACD";
char *str2 = "ACDA";
if (areRotations(str1, str2))
printf("Strings are rotations of each other");
else
printf("Strings are not rotations of each other");
getchar();
return 0;
}
Java
// Java program to check if two given strings are rotations of
// each other
class StringRotation
{
/* Function checks if passed strings (str1 and str2)
are rotations of each other */
static boolean areRotations(String str1, String str2)
{
// There lengths must be same and str2 must be
// a substring of str1 concatenated with str1.
return (str1.length() == str2.length()) &&
((str1 + str1).indexOf(str2) != -1);
}
// Driver method
public static void main (String[] args)
{
String str1 = "AACD";
String str2 = "ACDA";
if (areRotations(str1, str2))
System.out.println("Strings are rotations of each other");
else
System.out.printf("Strings are not rotations of each other");
}
}
// This code is contributed by munjal
Python3
# Python program to check if strings are rotations of
# each other or not
# Function checks if passed strings (str1 and str2)
# are rotations of each other
def areRotations(string1, string2):
size1 = len(string1)
size2 = len(string2)
temp = ''
# Check if sizes of two strings are same
if size1 != size2:
return 0
# Create a temp string with value str1.str1
temp = string1 + string1
# Now check if str2 is a substring of temp
# string.count returns the number of occurrences of
# the second string in temp
if (temp.count(string2)> 0):
return 1
else:
return 0
# Driver program to test the above function
string1 = "AACD"
string2 = "ACDA"
if areRotations(string1, string2):
print ("Strings are rotations of each other")
else:
print ("Strings are not rotations of each other")
# This code is contributed by Bhavya Jain
C#
// C# program to check if two given strings
// are rotations of each other
using System;
class GFG {
/* Function checks if passed strings
(str1 and str2) are rotations of
each other */
static bool areRotations(String str1,
String str2)
{
// There lengths must be same and
// str2 must be a substring of
// str1 concatenated with str1.
return (str1.Length == str2.Length )
&& ((str1 + str1).IndexOf(str2)
!= -1);
}
// Driver method
public static void Main ()
{
String str1 = "FGABCDE";
String str2 = "ABCDEFG";
if (areRotations(str1, str2))
Console.Write("Strings are"
+ " rotation s of each other");
else
Console.Write("Strings are "
+ "not rotations of each other");
}
}
// This code is contributed by nitin mittal.
PHP
<?php
// Php program to check if
// two given strings are
// rotations of each other
/* Function checks if passed
strings (str1 and str2) are
rotations of each other */
function areRotations($str1, $str2)
{
/* Check if sizes of two
strings are same */
if (strlen($str1) != strlen($str2))
{
return false;
}
$temp = $str1 + $str1;
if ($temp.count($str2)> 0)
{
return true;
}
else
{
return false;
}
}
// Driver code
$str1 = "AACD";
$str2 = "ACDA";
if (areRotations($str1, $str2))
{
echo "Strings are rotations ".
"of each other";
}
else
{
echo "Strings are not " .
"rotations of each other" ;
}
// This code is contributed
// by Shivi_Aggarwal.
?>
Javascript
<script>
// javascript program to check if two given strings are rotations of
// each other
/* Function checks if passed strings (str1 and str2)
are rotations of each other */
function areRotations( str1, str2)
{
// There lengths must be same and str2 must be
// a substring of str1 concatenated with str1.
return (str1.length == str2.length) &&
((str1 + str1).indexOf(str2) != -1);
}
// Driver method
var str1 = "AACD";
var str2 = "ACDA";
if (areRotations(str1, str2))
document.write("Strings are rotations of each other");
else
document.write("Strings are not rotations of each other");
// This code is contributed by umadevi9616
</script>
C++
#include <bits/stdc++.h>
using namespace std;
bool check_rotation(string s, string goal)
{
if (s.size() != goal.size())
return false;
queue<char> q1;
for (int i = 0; i < s.size(); i++) {
q1.push(s[i]);
}
queue<char> q2;
for (int i = 0; i < goal.size(); i++) {
q2.push(goal[i]);
}
int k = goal.size();
while (k--) {
char ch = q2.front();
q2.pop();
q2.push(ch);
if (q2 == q1)
return true;
}
return false;
}
int main()
{
string s1 = "ABCD";
string s2 = "CDAB";
if (check_rotation(s1, s2))
cout << s2 << " is a rotated form of " << s1
<< endl;
else
cout << s2 << " is not a rotated form of " << s1
<< endl;
string s3 = "ACBD";
if (check_rotation(s1, s3))
cout << s3 << " is a rotated form of " << s1
<< endl;
else
cout << s3 << " is not a rotated form of " << s1
<< endl;
return 0;
}
//CDAB
//ACBD
Java
import java.util.*;
class GFG{
static boolean check_rotation(String s, String goal)
{
if (s.length() != goal.length())
;
Queue<Character> q1 = new LinkedList<>();
for (int i = 0; i < s.length(); i++) {
q1.add(s.charAt(i));
}
Queue<Character> q2 = new LinkedList<>();
for (int i = 0; i < goal.length(); i++) {
q2.add(goal.charAt(i));
}
int k = goal.length();
while (k>0) {
k--;
char ch = q2.peek();
q2.remove();
q2.add(ch);
if (q2.equals(q1))
return true;
}
return false;
}
public static void main(String[] args)
{
String s1 = "ABCD";
String s2 = "CDAB";
if (check_rotation(s1, s2))
System.out.print(s2+ " is a rotated form of " + s1
+"\n");
else
System.out.print(s2+ " is not a rotated form of " + s1
+"\n");
String s3 = "ACBD";
if (check_rotation(s1, s3))
System.out.print(s3+ " is a rotated form of " + s1
+"\n");
else
System.out.print(s3+ " is not a rotated form of " + s1
+"\n");
}
}
// This code is contributed by gauravrajput1
Python3
def check_rotation(s, goal): if (len(s) != len(goal)): skip q1 = [] for i in range(len(s)): q1.insert(0, s[i]) q2 = [] for i in range(len(goal)): q2.insert(0, goal[i]) k = len(goal) while (k > 0): ch = q2[0] q2.pop(0) q2.append(ch) if (q2 == q1): return True k -= 1 return False if __name__ == "__main__": s1 = "ABCD" s2 = "CDAB" if (check_rotation(s1, s2)): print(s2, " is a rotated form of ", s1) else: print(s2, " is not a rotated form of ", s1) s3 = "ACBD" if (check_rotation(s1, s3)): print(s3, " is a rotated form of ", s1) else: print(s3, " is not a rotated form of ", s1) # This code is contributed by ukasp.
C#
using System;
using System.Collections;
public class GFG{
public static bool check_rotation(string s, string goal)
{
if (s.Length != goal.Length)
return false;
Queue q1 = new Queue();
for (int i = 0; i < s.Length; i++) {
q1.Enqueue(s[i]);
}
Queue q2 = new Queue();
for (int i = 0; i < goal.Length; i++) {
q2.Enqueue(goal[i]);
}
int k = goal.Length;
for(int i=k-1;i>=0;i--) {
char ch = (char)(q2.Peek());
q2.Dequeue();
q2.Enqueue(ch);
if (q2.Equals(q1))
return true;
}
return false;
}
static public void Main (){
string s1 = "ABCD";
string s2 = "CDAB";
if (!check_rotation(s1, s2))
{ Console.Write(s2);
Console.Write(" is a rotated form of ");
Console.WriteLine(s1);
}
else
{
Console.Write(s2);
Console.Write(" is not a rotated form of ");
Console.WriteLine(s1);
}
string s3 = "ACBD";
if (check_rotation(s1, s3))
{ Console.Write(s3);
Console.Write(" is a rotated form of ");
Console.WriteLine(s1);
}
else
{
Console.Write(s3);
Console.Write(" is not a rotated form of ");
Console.WriteLine(s1);
}
}
}
// This code is contributed by akashish_.
Javascript
<script>
function check_rotation(s, goal){
if (s.length != goal.length){
return false;
}
let q1 = []
for(let i=0;i<s.length;i++)
q1.push(s[i])
let q2 = []
for(let i=0;i<goal.length;i++)
q2.push(goal[i])
let k = goal.length
while (k--){
let ch = q2[0]
q2.shift()
q2.push(ch)
if (JSON.stringify(q2) == JSON.stringify(q1))
return true
}
return false
}
// driver code
let s1 = "ABCD"
let s2 = "CDAB"
if (check_rotation(s1, s2))
document.write(s2, " is a rotated form of ", s1,"</br>")
else
document.write(s2, " is not a rotated form of ", s1,"</br>")
let s3 = "ACBD"
if (check_rotation(s1, s3))
document.write(s3, " is a rotated form of ", s1,"</br>")
else
document.write(s3, " is not a rotated form of ", s1,"</br>")
// This code is contributed by shinjanpatra.
</script>
C++
#include <iostream>
#include <vector>
using namespace std;
bool checkString(string &s1, string &s2, int indexFound, int Size)
{
for(int i=0;i<Size;i++){
//check whether the character is equal or not
if(s1[i]!=s2[(indexFound+i)%Size])return false;
// %Size keeps (indexFound+i) in bounds, since it
// ensures it's value is always less than Size
}
return true;
}
int main() {
string s1="abcd";
string s2="cdab";
if(s1.length()!=s2.length()){
cout<<"s2 is not a rotation on s1"<<endl;
}
else{
// store occurrences of the first character of s1
vector<int> indexes;
int Size = s1.length();
char firstChar = s1[0];
for(int i=0;i<Size;i++)
{
if(s2[i]==firstChar)
{
indexes.push_back(i);
}
}
bool isRotation=false;
// check if the strings are rotation of each other for
// every occurence of firstChar in s2
for(int idx: indexes)
{
isRotation = checkString( s1, s2, idx, Size);
if(isRotation)
break;
}
if(isRotation)cout<<"s2 is rotation of s1"<<endl;
else cout<<"s2 is not a rotation of s1"<<endl;
}
return 0;
}
Java
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG
{
// java program to check if two strings are rotation of each other or not
static boolean checkString(String s1, String s2, int indexFound, int Size)
{
for(int i=0;i<Size;i++)
{
//check whether the character is equal or not
if(s1.charAt(i) != s2.charAt((indexFound+i)%Size))return false;
// %Size keeps (indexFound+i) in bounds,
// since it ensures it's value is always less than Size
}
return true;
}
// Driver code
public static void main(String args[])
{
String s1="abcd";
String s2="cdab";
if(s1.length() != s2.length()){
System.out.println("s2 is not a rotation on s1");
}
else{
ArrayList<Integer>indexes = new ArrayList<Integer>(); //store occurrences of the first character of s1
int Size = s1.length();
char firstChar = s1.charAt(0);
for(int i=0;i<Size;i++)
{
if(s2.charAt(i)==firstChar)
{
indexes.add(i);
}
}
boolean isRotation=false;
// check if the strings are rotation of each other for every occurence of firstChar in s2
for(int idx: indexes)
{
isRotation = checkString(s1, s2, idx, Size);
if(isRotation)
break;
}
if(isRotation)System.out.println("s2 is rotation of s1");
else System.out.println("s2 is not a rotation of s1");
}
}
}
// This code is contributed by shinjanpatra
Python3
def checkString(s1, s2, indexFound, Size):
for i in range(Size):
# check whether the character is equal or not
if(s1[i] != s2[(indexFound + i) % Size]):
return False
# %Size keeps (indexFound+i) in bounds,
# since it ensures it's value is always less than Size
return True
# driver code
s1 = "abcd"
s2 = "cdab"
if(len(s1) != len(s2)):
print("s2 is not a rotation on s1")
else:
indexes = [] #store occurrences of the first character of s1
Size = len(s1)
firstChar = s1[0]
for i in range(Size):
if(s2[i] == firstChar):
indexes.append(i)
isRotation = False
# check if the strings are rotation of each other
# for every occurence of firstChar in s2
for idx in indexes:
isRotation = checkString(s1, s2, idx, Size)
if(isRotation):
break
if(isRotation):
print("s2 is rotation of s1")
else:
print("s2 is not a rotation of s1")
# This code is contributed by shinjanpatra
C#
using System;
public class GFG{
public static bool checkString(string s1, string s2, int indexFound, int Size)
{
for(int i = 0; i < Size; i++)
{
//check whether the character is equal or not
if(s1[i] != s2[(indexFound+i)%Size])
return false;
// %Size keeps (indexFound+i) in bounds, since it
// ensures it's value is always less than Size
}
return true;
}
static public void Main (){
string s1="abcd";
string s2="cdab";
if(s1.Length != s2.Length){
Console.WriteLine("s2 is not a rotation on s1");
}
else
{
// store occurrences of the first character of s1
int[] indexes = new int[1000];
int j = 0;
int Size = s1.Length;
char firstChar = s1[0];
for(int i = 0; i < Size; i++)
{
if(s2[i] == firstChar)
{
indexes[j] = i;
j++;
}
}
bool isRotation=false;
// check if the strings are rotation of each other for
// every occurence of firstChar in s2
for(int idx = 0; idx < indexes.Length; idx++)
{
isRotation = checkString( s1, s2, idx, Size);
if(isRotation)
break;
}
if(isRotation)
Console.WriteLine("s2 is rotation of s1");
else
Console.WriteLine("s2 is not a rotation of s1");
}
}
}
// This code is contributed by akashish__
Javascript
<script>
function checkString(s1, s2, indexFound, Size)
{
for(let i = 0; i < Size; i++)
{
//check whether the character is equal or not
if(s1[i] != s2[(indexFound + i) % Size])return false;
// %Size keeps (indexFound+i) in bounds, since it ensures it's value is always less than Size
}
return true;
}
// driver code
let s1 = "abcd";
let s2 = "cdab";
if(s1.length != s2.length)
{
document.write("s2 is not a rotation on s1");
}
else
{
let indexes = []; //store occurrences of the first character of s1
let Size = s1.length;
let firstChar = s1[0];
for(let i = 0; i < Size; i++)
{
if(s2[i] == firstChar)
{
indexes.push(i);
}
}
let isRotation = false;
// check if the strings are rotation of each other for every occurence of firstChar in s2
for(let idx of indexes)
{
isRotation = checkString(s1, s2, idx, Size);
if(isRotation)
break;
}
if(isRotation)document.write("s2 is rotation of s1")
else document.write("s2 is not a rotation of s1")
}
// This code is contributed by shinjanpatra
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA