检查字符串是否彼此旋转。套装2

给定两个字符串s1和s2，请检查s2是否为s1的旋转。
例子：

Input : ABACD, CDABA
Output : True

Input : GEEKS, EKSGE
Output : True

我们已经在较早的文章中讨论了一种将子字符串匹配作为一种模式进行处理的方法。在这篇文章中，我们将要使用KMP算法的LPS(最长适当的前缀，这也是后缀)建设，这将在找到字符串B和字符串的后缀的前缀的最长匹配的帮助。通过它我们将知道旋转点，从这个点匹配字符。如果所有字符都匹配，则为旋转，否则为非。
以下是上述方法的基本实现。

C++

// C++ program to check if
// two strings are rotations
// of each other
#include
using namespace std;
bool isRotation(string a,
                string b)
{
  int n = a.length();
  int m = b.length();
  if (n != m)
    return false;
 
  // create lps[] that
  // will hold the longest
  // prefix suffix values
  // for pattern
  int lps[n];
 
  // length of the previous
  // longest prefix suffix
  int len = 0;
  int i = 1;
   
  // lps[0] is always 0
  lps[0] = 0;
 
  // the loop calculates
  // lps[i] for i = 1 to n-1
  while (i < n)
  {
    if (a[i] == b[len])
    {
      lps[i] = ++len;
      ++i;
    }
    else
    {
      if (len == 0)
      {
        lps[i] = 0;
        ++i;
      }
      else
      {
        len = lps[len - 1];
      }
    }
  }
 
  i = 0;
 
  // Match from that rotating
  // point
  for (int k = lps[n - 1];
           k < m; ++k)
  {
    if (b[k] != a[i++])
      return false;
  }
  return true;
}
 
// Driver code
int main()
{
  string s1 = "ABACD";
  string s2 = "CDABA";
  cout << (isRotation(s1, s2) ?
           "1" : "0");
}
 
// This code is contributed by Chitranayal

Java

// Java program to check if two strings are rotations
// of each other.
import java.util.*;
import java.lang.*;
import java.io.*;
class stringMatching {
    public static boolean isRotation(String a, String b)
    {
        int n = a.length();
        int m = b.length();
        if (n != m)
            return false;
 
        // create lps[] that will hold the longest
        // prefix suffix values for pattern
        int lps[] = new int[n];
 
        // length of the previous longest prefix suffix
        int len = 0;
        int i = 1;
        lps[0] = 0; // lps[0] is always 0
 
        // the loop calculates lps[i] for i = 1 to n-1
        while (i < n) {
            if (a.charAt(i) == b.charAt(len)) {
                lps[i] = ++len;
                ++i;
            }
            else {
                if (len == 0) {
                    lps[i] = 0;
                    ++i;
                }
                else {
                    len = lps[len - 1];
                }
            }
        }
 
        i = 0;
 
        // match from that rotating point
        for (int k = lps[n - 1]; k < m; ++k) {
            if (b.charAt(k) != a.charAt(i++))
                return false;
        }
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s1 = "ABACD";
        String s2 = "CDABA";
 
        System.out.println(isRotation(s1, s2) ? "1" : "0");
    }
}

Python3

# Python program to check if
# two strings are rotations
# of each other
def isRotation(a: str, b: str) -> bool:
    n = len(a)
    m = len(b)
    if (n != m):
        return False
 
    # create lps[] that
    # will hold the longest
    # prefix suffix values
    # for pattern
    lps = [0 for _ in range(n)]
 
    # length of the previous
    # longest prefix suffix
    length = 0
    i = 1
 
    # lps[0] is always 0
    lps[0] = 0
 
    # the loop calculates
    # lps[i] for i = 1 to n-1
    while (i < n):
        if (a[i] == b[length]):
            length += 1
            lps[i] = length
            i += 1
        else:
            if (length == 0):
                lps[i] = 0
                i += 1
            else:
                length = lps[length - 1]
    i = 0
 
    # Match from that rotating
    # point
    for k in range(lps[n - 1], m):
        if (b[k] != a[i]):
            return False
        i += 1
    return True
 
# Driver code
if __name__ == "__main__":
 
    s1 = "ABACD"
    s2 = "CDABA"
    print("1" if isRotation(s1, s2) else "0")
 
# This code is contributed by sanjeev2552

C#

// C# program to check if
// two strings are rotations
// of each other.
using System;
 
class GFG
{
public static bool isRotation(string a,
                              string b)
{
    int n = a.Length;
    int m = b.Length;
    if (n != m)
        return false;
 
    // create lps[] that will
    // hold the longest prefix
    // suffix values for pattern
    int []lps = new int[n];
 
    // length of the previous
    // longest prefix suffix
    int len = 0;
    int i = 1;
     
    // lps[0] is always 0
    lps[0] = 0;
 
    // the loop calculates
    // lps[i] for i = 1 to n-1
    while (i < n)
    {
        if (a[i] == b[len])
        {
            lps[i] = ++len;
            ++i;
        }
        else
        {
            if (len == 0)
            {
                lps[i] = 0;
                ++i;
            }
            else
            {
                len = lps[len - 1];
            }
        }
    }
 
    i = 0;
 
    // match from that
    // rotating point
    for (int k = lps[n - 1]; k < m; ++k)
    {
        if (b[k] != a[i++])
            return false;
    }
    return true;
}
 
// Driver code
public static void Main()
{
    string s1 = "ABACD";
    string s2 = "CDABA";
 
    Console.WriteLine(isRotation(s1, s2) ?
                                     "1" : "0");
}
}
 
// This code is contributed
// by anuj_67.

Javascript

 
// javascript program to check if two strings are rotations
// of each other.
function isRotation(a, b)
{
    var n = a.length;
    var m = b.length;
    if (n != m)
        return false;
 
    // create lps that will hold the longest
    // prefix suffix values for pattern
    var lps = Array.from({length: n}, (_, i) => 0);
 
    // length of the previous longest prefix suffix
    var len = 0;
    var i = 1;
    lps[0] = 0; // lps[0] is always 0
 
    // the loop calculates lps[i] for i = 1 to n-1
    while (i < n) {
        if (a.charAt(i) == b.charAt(len)) {
            lps[i] = ++len;
            ++i;
        }
        else {
            if (len == 0) {
                lps[i] = 0;
                ++i;
            }
            else {
                len = lps[len - 1];
            }
        }
    }
 
    i = 0;
 
    // match from that rotating point
    for (k = lps[n - 1]; k < m; ++k) {
        if (b.charAt(k) != a.charAt(i++))
            return false;
    }
    return true;
}
 
// Driver code
var s1 = "ABACD";
var s2 = "CDABA";
document.write(isRotation(s1, s2) ? "1" : "0");
 
// This code is contributed by shikhasingrajput.

输出：

1

时间复杂度： O(n)
辅助空间： O(n)