检查是否可以通过多次反转子数组来使两个数组相等

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数来检查是否可以使数组 B 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：由于我们必须通过反转任何子数组任意次数才能使数组B等于数组A，因此只有当两个数组都是字谜时才有可能。要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是否有任何索引元素不相同，然后我们将返回 false，否则我们将在最后返回 true。

下面是上述方法的实现：

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

Javascript

 
 
// Javascript implementation to check if
// two arrays can be made equal
 
// Function to check if array B
// can be made equal to array A
function canMadeEqual(A, B, n)
{
    // sort both the arrays
    A.sort();
    B.sort();
 
    // Check if both the arrays
    // are equal or not
    for (var i = 0; i < n; i++)
        if (A[i] != B[i])
            return false;
    return true;
}
 
// Driver Code
var A = [ 1, 2, 3 ];
var B = [ 1, 3, 2 ];
var n = A.length;
if (canMadeEqual(A, B, n))
    document.write( "Yes");
else
    document.write("No");
 
// This code is contributed by rutvik_56.

输出

Yes

时间复杂度： O(N*Log N)
辅助空间： O(1)

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