查找N个字符串的最大公约数(GCD)的程序

给定一个字符串数组arr [] ，该任务是给定的字符串数组的最大公约数。

In strings ‘A’ and ‘B’, we say “B divides A” if and only if A = concatenation of B more than 1 times.Find the largest string which divides both A and B. 
 

例子：

Input: arr[] = { “GFGGFG”, “GFGGFG”, “GFGGFGGFGGFG” } 
Output: “GFGGFG” 
Explanation: 
“GFGGFG” is the largest string which divides the whole array elements.
Input: arr = { “Geeks”, = “GFG”} 
Output: “” 

方法：想法是使用递归。步骤如下：

1. 创建一个递归函数gcd(str1，str2) 。
2. 如果str2的长度大于str1，那么我们将使用gcd(str2，str1)进行递归。
3. 现在，如果str1不是以str2开头，则返回一个空字符串。
4. 如果更长的字符串用较短的字符串开始，切断长字符串并易复发或重复的共同前缀部分，直到一个是空的。
5. 在上述步骤之后返回的字符串是给定字符串数组的gcd。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include
using namespace std;
 
// Function that finds gcd of 2 strings
string gcd(string str1, string str2)
{
   
    // If str1 length is less than
    // that of str2 then recur
    // with gcd(str2, str1)
    if (str1.length() < str2.length())
    {
        return gcd(str2, str1);
    }
   
    // If str1 is not the
    // concatenation of str2
    else if(str1.find(str2) != 0)
    {
        return "";
    }
    else if (str2 == "")
    {
       
        // GCD string is found
        return str1;
    }
    else
    {
       
        // Cut off the common prefix
        // part of str1 & then recur
        return gcd(str1.substr(str2.length()), str2);
    }
}
 
// Function to find GCD of array of
// strings
string findGCD(string arr[], int n)
{
    string result = arr[0];
    for (int i = 1; i < n; i++)
    {
        result = gcd(result, arr[i]);
    }
   
    // Return the GCD of strings
    return result;
}
 
// Driver Code
int main()
{
   
    // Given array  of strings
    string arr[]={ "GFGGFG",
                         "GFGGFG",
                         "GFGGFGGFGGFG" };
    int n = sizeof(arr)/sizeof(arr[0]);
   
    // Function Call
    cout << findGCD(arr, n);
}
 
// This code is contributed by pratham76.

Java

// Java program for the above approach
 
class GCD {
 
    // Function that finds gcd of 2 strings
    static String gcd(String str1, String str2)
    {
        // If str1 length is less than
        // that of str2 then recur
        // with gcd(str2, str1)
        if (str1.length() < str2.length()) {
            return gcd(str2, str1);
        }
 
        // If str1 is not the
        // concatenation of str2
        else if (!str1.startsWith(str2)) {
            return "";
        }
 
        else if (str2.isEmpty()) {
 
            // GCD string is found
            return str1;
        }
        else {
 
            // Cut off the common prefix
            // part of str1 & then recur
            return gcd(str1.substring(str2.length()),
                       str2);
        }
    }
 
    // Function to find GCD of array of
    // strings
    static String findGCD(String arr[], int n)
    {
        String result = arr[0];
 
        for (int i = 1; i < n; i++) {
            result = gcd(result, arr[i]);
        }
 
        // Return the GCD of strings
        return result;
    }
 
    // Driver Code
    public static void
        main(String[] args)
    {
        // Given array  of strings
        String arr[]
            = new String[] { "GFGGFG",
                             "GFGGFG",
                             "GFGGFGGFGGFG" };
        int n = arr.length;
 
        // Function Call
        System.out.println(findGCD(arr, n));
    }
}

Python3

# Python3 program for the above approach
 
# Function that finds gcd of 2 strings
def gcd(str1, str2):
 
    # If str1 length is less than
    # that of str2 then recur
    # with gcd(str2, str1)
    if(len(str1) < len(str2)):
        return gcd(str2, str1)
     
    # If str1 is not the
    # concatenation of str2
    elif(not str1.startswith(str2)):
        return ""
    elif(len(str2) == 0):
         
        # GCD string is found
        return str1
    else:
         
        # Cut off the common prefix
        # part of str1 & then recur
        return gcd(str1[len(str2):], str2)
 
# Function to find GCD of array of
# strings
def findGCD(arr, n):
    result = arr[0]
 
    for i in range(1, n):
        result = gcd(result, arr[i])
     
    # Return the GCD of strings
    return result
 
# Driver Code
 
# Given array  of strings
arr = ["GFGGFG", "GFGGFG", "GFGGFGGFGGFG" ]
n = len(arr)
 
# Function Call
print(findGCD(arr, n))
 
# This code is contributed by avanitrachhadiya2155

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function that finds gcd
// of 2 strings
static String gcd(String str1,
                  String str2)
{
  // If str1 length is less than
  // that of str2 then recur
  // with gcd(str2, str1)
  if (str1.Length < str2.Length)
  {
    return gcd(str2, str1);
  }
 
  // If str1 is not the
  // concatenation of str2
  else if (!str1.StartsWith(str2))
  {
    return "";
  }
  else if (str2.Length == 0)
  {
    // GCD string is found
    return str1;
  }
  else
  {
    // Cut off the common prefix
    // part of str1 & then recur
    return gcd(str1.Substring(str2.Length),
                              str2);
  }
}
 
// Function to find GCD
// of array of strings
static String findGCD(String []arr,
                      int n)
{
  String result = arr[0];
 
  for (int i = 1; i < n; i++)
  {
    result = gcd(result, arr[i]);
  }
 
  // Return the GCD of strings
  return result;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array  of strings
  String []arr = new String[] {"GFGGFG",
                               "GFGGFG",
                               "GFGGFGGFGGFG"};
  int n = arr.Length;
 
  // Function Call
  Console.WriteLine(findGCD(arr, n));
}
}
 
// This code is contributed by shikhasingrajput

输出：

GFGGFG

时间复杂度： O(N * log(B))，其中N是字符串的数量，而B是arr []中任何字符串的最大长度。
辅助空间： O(1)