通过多次反转子数组来检查两个数组是否相等

给定两个数组A []和B []，任务是通过将B的子数组反转任意次来检查数组B是否可以等于A。

例子：

Input:  A[] =  {1 2 3}, B[] =  {3 1 2}
Output: Yes
Explanation: 
Reverse subarrays in array B as shown below:
Reverse subarray [3, 1], B becomes [1, 3, 2]
Reverse subarray [3, 2], B becomes [1, 2, 3] = A
There are multiple ways to convert B to A.

Input:  A[] = {1 2 3}, B[] = {3 4 1 }
Output: No
 

方法：由于只需要通过反转任意次数组数次就可以使数组B与数组A相等，因此只有当两个数组都为字谜时才有可能。为了检查两个数组是否都是字谜，我们将对两个数组进行排序，并检查在任何索引元素上是否不相同，然后我们将返回false，否则最后将返回true。

下面实现以上方法：

C++

// C++ implementation to check if 
// two arrays can be made equal
  
#include 
using namespace std;
  
// Function to check if array B
// can be made equal to array A
bool canMadeEqual(int A[], 
                  int B[], int n)
{
    // sort both the arrays
    sort(A, A + n);
    sort(B, B + n);
  
    // Check if both the arrays
    // are equal or not
    for (int i = 0; i < n; i++)
        if (A[i] != B[i])
            return false;
    return true;
}
  
// Driver Code
int main()
{
    int A[] = { 1, 2, 3 };
    int B[] = { 1, 3, 2 };
    int n = sizeof(A) / sizeof(A[0]);
    if (canMadeEqual(A, B, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

C

// C implementation to check if  
// two arrays can be made equal 
#include
#include
  
int sort(int a[],int n)
{
    int i, j, tmp;
    for(i = 0; i < n; i++)
    {
        for(j = i + 1; j < n; j++)
        {
            if(a[j] 

---

Java

// Java implementation to check if 
// two arrays can be made equal 
import java.util.*;
  
class GFG{
      
// Function to check if array B 
// can be made equal to array A 
public static boolean canMadeEqual(int[] A,
                                   int[] B,
                                   int n) 
{ 
      
    // Sort both the arrays 
    Arrays.sort(A); 
    Arrays.sort(B); 
  
    // Check if both the arrays 
    // are equal or not 
    for(int i = 0; i < n; i++) 
    {
        if (A[i] != B[i]) 
        {
            return false; 
        }
    }
    return true; 
}
  
// Driver code
public static void main(String[] args)
{
    int A[] = { 1, 2, 3 }; 
    int B[] = { 1, 3, 2 }; 
    int n = A.length; 
      
    if (canMadeEqual(A, B, n)) 
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
  
// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to check if 
# two arrays can be made equal
  
# Function to check if array B
# can be made equal to array A
def canMadeEqual(A, B, n):
    
  # Sort both the arrays
  A.sort()
  B.sort()
  
  # Check if both the arrays
  # are equal or not
  for i in range(n):
    if (A[i] != B[i]):
      return False
      
  return True
  
# Driver Code
if __name__ == "__main__":
    
  A = [ 1, 2, 3 ]
  B = [ 1, 3, 2 ]
  n = len(A)
  
  if (canMadeEqual(A, B, n)):
    print( "Yes")
  else:
    print("No")
  
# This code is contributed by chitranayal

C#

// C# implementation to check if 
// two arrays can be made equal 
using System;
  
class GFG{ 
  
// Function to check if array B 
// can be made equal to array A 
static bool canMadeEqual(int[] A, 
                         int[] B,
                         int n) 
{ 
      
    // Sort both the arrays 
    Array.Sort(A); 
    Array.Sort(B); 
  
    // Check if both the arrays 
    // are equal or not 
    for(int i = 0; i < n; i++) 
    {
        if (A[i] != B[i]) 
        {
            return false; 
        }
    }
    return true; 
} 
  
// Driver code 
public static void Main() 
{ 
    int[] A = { 1, 2, 3 }; 
    int[] B = { 1, 3, 2 }; 
    int n = A.Length; 
      
    if (canMadeEqual(A, B, n)) 
        Console.Write("Yes");
    else
        Console.Write("No"); 
} 
} 
  
// This code is contributed by adityakumar27200

输出

Yes

时间复杂度： O(N * Log N)
辅助空间： O(1)