通过多次反转子阵列检查两个阵列是否可以相等
原文:https://www . geeksforgeeks . org/check-if-two-array-可通过多次反转子阵列使其相等/
给定两个阵列 A[] 和 B[]，，任务是通过将 B 的子阵列反转任意次数来检查阵列 B 是否可以等于 A。
示例:
输入: A[] = {1 2 3}，B[] = {3 1 2} 输出:是 解释: 反向子阵在阵列 B 中如下所示: 反向子阵【3，1】，B 变成【1，3，2】 反向子阵【3，2】，B 变成【1，2，3】= A 有多种方式将 B 转换为 A. 输入:
逼近:由于我们只需要将任意一个子数组反转任意多次就可以使数组 B 等于数组 A，所以只有当两个数组都是字谜时才有可能。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任何索引元素是否不相同，然后我们将返回 false，否则我们将在最后返回 true。
以下实施上述方法:
C++
// C++ implementation to check if
// two arrays can be made equal
#include <bits/stdc++.h>
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<stdio.h>
#include<math.h>
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] <a[i])
{
tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
}
}
return 0;
}
// Function to check if array B
// can be made equal to array A
int canMadeEqual(int A[], int B[], int n)
{
int i;
// Sort both arrays
sort(A, n);
sort(B, n);
// Check both arrays equal or not
for(i = 0; i < n; i++)
{
if (A[i] != B[i])
{
return(0);
}
}
return (1);
}
// Driver Code
int main()
{
int A[] = { 1, 2, 3 };
int n;
int B[] = { 1, 3, 2 };
n = sizeof(A) / sizeof(A[0]);
if(canMadeEqual(A, B, n))
{
printf("Yes");
}
else
{
printf("No");
}
return 0;
}
// This code is contributed by adityakumar27200
Java 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// 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.
</script>
Output
Yes
时间复杂度:O(N * Log N) T5】辅助空间: O(1)
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