检查字符串是否相互旋转的程序

给定一个字符串s1 和一个字符串s2，写一个片段来说明 s2 是否是 s1 的旋转？
（例如，给定 s1 = ABCD 和 s2 = CDAB，返回真，给定 s1 = ABCD，和 s2 = ACBD，返回假）

算法： areRotations(str1, str2)

1. Create a temp string and store concatenation of str1 to
       str1 in temp.
                          temp = str1.str1
    2. If str2 is a substring of temp then str1 and str2 are 
       rotations of each other.

    Example:                 
                     str1 = "ABACD"
                     str2 = "CDABA"

     temp = str1.str1 = "ABACDABACD"
     Since str2 is a substring of temp, str1 and str2 are 
     rotations of each other.

C++

// C++ program to check if two given strings
// are rotations of  each other
# include 
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 
# include 
# include 
 
/* 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

 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

// 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

---

输出

Strings are rotations of each other

方法 2（使用 STL）：

算法 ：

1.如果两个字符串的大小不相等，那么永远不可能。

2. 将原始字符串推入队列q1 。

3. 将要检查的字符串推入另一个队列q2中。

4. 继续弹出q2并将其推回其中，直到此类操作的数量小于字符串的大小。

5. 如果在这些操作期间的任何时候q2等于q1 ，这是可能的。否则没有。

C++

#include 
using namespace std;
bool check_rotation(string s, string goal)
{
    if (s.size() != goal.size())
        ;
    queue q1;
    for (int i = 0; i < s.size(); i++) {
        q1.push(s[i]);
    }
    queue 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;
}

Java

import java.util.*;
 
class GFG{
static boolean check_rotation(String s, String goal)
{
    if (s.length() != goal.length())
        ;
    Queue q1 = new LinkedList<>();
    for (int i = 0; i < s.length(); i++) {
        q1.add(s.charAt(i));
    }
    Queue 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.

Javascript

 
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.
 

输出

CDAB is a rotated form of ABCD
ACBD is not a rotated form of ABCD

使用的库函数：
字符串：
strstr 在 string 中查找子字符串。
原型：char * strstr(const char *s1, const char *s2);
有关详细信息，请参阅 http://www.lix.polytechnique.fr/Labo/Leo.Liberti/public/computing/prog/c/C/MAN/strstr.htm
字符串：
strncat 连接两个字符串
原型：char *strcat(char *dest, const char *src);
有关详细信息，请参阅 http://www.lix.polytechnique.fr/Labo/Leo.Liberti/public/computing/prog/c/C/MAN/strcat.htm

时间复杂度：这个问题的时间复杂度取决于 strstr函数的实现。
如果 strstr 的实现是使用 KMP 匹配器完成的，那么上述程序的复杂度是 (-)(n1 + n2) 其中 n1 和 n2 是字符串的长度。 KMP 匹配器花费 (-)(n) 时间在长度为 n 的字符串中查找子字符串，其中假设子字符串的长度小于字符串。

方法三：

算法：

1.查找待检查字符串中原字符串第一个字符的所有位置。

2.对于找到的每个位置，将其视为要检查的字符串的起始索引。

3.从新的起始索引开始，比较两个字符串并检查它们是否相等。

（假设原始字符串为s1 ，待检查字符串为s2 ，n为字符串长度， j为 s1 的第一个字符在 s2 中的位置，

然后对于 i < (原始字符串的长度)，检查是否 s1[i]==s2[(j+1)%n)。如果发现任何字符不匹配，则返回 false，否则返回 true。

4.对找到的所有位置重复第 3 步。

C++

#include 
#include 
using namespace std;
 
bool checkString(string &s1, string &s2, int indexFound, int Size)
{
    for(int i=0;i indexes; //store occurences of the first character of s1
       
      int Size = s1.length();
       
      char firstChar = s1[0];
     
     
      for(int i=0;i

Javascript

 
 
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 occurences 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
 

输出

s2 is rotation of s1

时间复杂度：

在最坏的情况下，时间复杂度将是 n*n，其中 n 是字符串的长度。