排列是将序列成员重新排列为新序列。例如，有[a，b，c，d]的24个排列。其中一些是[b，a，d，c] ， [d，a，b，c]和[a，d，b，c] 。
排列可以由数组P []指定，其中P [i]表示元素在排列中索引i处的位置。
例如，数组[ 3，2，1，0 ]表示将索引0的元素映射到索引3，将索引1的元素映射到索引2，将索引2的元素映射到索引1和将索引的元素映射到的排列。 3到索引0。
给定N个元素的数组arr []和置换数组P [] ，任务是基于置换数组P []置换给定的数组arr [] 。
例子：
Input: arr[] = {1, 2, 3, 4}, P[] = {3, 2, 1, 0}
Output: 4 3 2 1
Input: arr[] = {11, 32, 3, 42}, P[] = {2, 3, 0, 1}
Output: 3 42 11 32
方法：每个置换可以由一组独立的置换来表示，每个置换都是循环的，即它通过固定的偏移量环绕所有元素。要找到并应用指示输入i的循环，只需继续前进(从i到P [i] )，直到我们回到i为止。完成当前循环后，找到另一个尚未应用的循环。要对此进行检查，请在应用P [i]后减去n 。这意味着，如果P [i]中的条目为负，则我们执行了相应的移动。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to permute the the given
// array based on the given conditions
int permute(int A[], int P[], int n)
{
// For each element of P
for (int i = 0; i < n; i++) {
int next = i;
// Check if it is already
// considered in cycle
while (P[next] >= 0) {
// Swap the current element according
// to the permutation in P
swap(A[i], A[P[next]]);
int temp = P[next];
// Subtract n from an entry in P
// to make it negative which indicates
// the corresponding move
// has been performed
P[next] -= n;
next = temp;
}
}
}
// Driver code
int main()
{
int A[] = { 5, 6, 7, 8 };
int P[] = { 3, 2, 1, 0 };
int n = sizeof(A) / sizeof(int);
permute(A, P, n);
// Print the new array after
// applying the permutation
for (int i = 0; i < n; i++)
cout << A[i] << " ";
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to permute the the given
// array based on the given conditions
static void permute(int A[], int P[], int n)
{
// For each element of P
for (int i = 0; i < n; i++)
{
int next = i;
// Check if it is already
// considered in cycle
while (P[next] >= 0)
{
// Swap the current element according
// to the permutation in P
swap(A, i, P[next]);
int temp = P[next];
// Subtract n from an entry in P
// to make it negative which indicates
// the corresponding move
// has been performed
P[next] -= n;
next = temp;
}
}
}
static int[] swap(int []arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver code
public static void main(String[] args)
{
int A[] = { 5, 6, 7, 8 };
int P[] = { 3, 2, 1, 0 };
int n = A.length;
permute(A, P, n);
// Print the new array after
// applying the permutation
for (int i = 0; i < n; i++)
System.out.print(A[i]+ " ");
}
}
// This code is contributed by 29AjayKumarPython 3
# Python 3 implementation of the approach
# Function to permute the the given
# array based on the given conditions
def permute(A, P, n):
# For each element of P
for i in range(n):
next = i
# Check if it is already
# considered in cycle
while (P[next] >= 0):
# Swap the current element according
# to the permutation in P
t = A[i]
A[i] = A[P[next]]
A[P[next]] = t
temp = P[next]
# Subtract n from an entry in P
# to make it negative which indicates
# the corresponding move
# has been performed
P[next] -= n
next = temp
# Driver code
if __name__ == '__main__':
A = [5, 6, 7, 8]
P = [3, 2, 1, 0]
n = len(A)
permute(A, P, n)
# Print the new array after
# applying the permutation
for i in range(n):
print(A[i], end = " ")
# This code is contributed by Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to permute the the given
// array based on the given conditions
static void permute(int []A, int []P, int n)
{
// For each element of P
for (int i = 0; i < n; i++)
{
int next = i;
// Check if it is already
// considered in cycle
while (P[next] >= 0)
{
// Swap the current element according
// to the permutation in P
swap(A, i, P[next]);
int temp = P[next];
// Subtract n from an entry in P
// to make it negative which indicates
// the corresponding move
// has been performed
P[next] -= n;
next = temp;
}
}
}
static int[] swap(int []arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver code
public static void Main()
{
int []A = { 5, 6, 7, 8 };
int []P = { 3, 2, 1, 0 };
int n = A.Length;
permute(A, P, n);
// Print the new array after
// applying the permutation
for (int i = 0; i < n; i++)
Console.Write(A[i]+ " ");
}
}
// This code is contributed by AnkitRai01输出：
8 7 6 5
时间复杂度： O(n)
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