逆排列
给定一个大小为 n 的整数数组，范围从 1 到 n，我们需要找到该数组的逆排列。 逆置换是通过在数组中元素值指定的位置插入元素的位置而得到的置换。为了更好地理解，考虑下面的例子: 假设我们在一个数组中的位置 3 找到元素 4，然后在反向排列中，我们在位置 4(元素值)插入 3(元素 4 在数组中的位置)。 基本上，逆置换是一种置换，其中每个数和它所占据的位置的数被交换。 数组应该包含从 1 到 array_size 的元素。 例 1 :
Input = {1, 4, 3, 2}
Output = {1, 4, 3, 2}
在这种情况下，对于元素 1，我们从 arr1 插入位置 1，即 arr2 中位置 1 的 1。对于 arr1 中的元素 4，我们在 arr2 的位置 4 插入 arr1 中的 2。类似地，对于 arr1 中的元素 2，我们在 arr2 中插入位置 2，即 4。
Example 2 :
Input = {2, 3, 4, 5, 1}
Output = {5, 1, 2, 3, 4}
在这个例子中，对于元素 2，我们在 arr2 的位置 2 插入 arr1 的位置 2。同样，我们发现其他元素的逆排列。 考虑一个数组 arr，数组 arr 中有元素 1 到 n. 方法 1 : 在这个方法中，我们一个一个地取元素，按递增的顺序检查元素，并打印出找到那个元素的元素的位置。
C++
// Naive CPP Program to find inverse permutation.
#include <bits/stdc++.h>
using namespace std;
// C++ function to find inverse permutations
void inversePermutation(int arr[], int size) {
// Loop to select Elements one by one
for (int i = 0; i < size; i++) {
// Loop to print position of element
// where we find an element
for (int j = 0; j < size; j++) {
// checking the element in increasing order
if (arr[j] == i + 1) {
// print position of element where
// element is in inverse permutation
cout << j + 1 << " ";
break;
}
}
}
}
// Driver program to test above function
int main() {
int arr[] = {2, 3, 4, 5, 1};
int size = sizeof(arr) / sizeof(arr[0]);
inversePermutation(arr, size);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Naive java Program to find inverse permutation.
import java.io.*;
class GFG {
// java function to find inverse permutations
static void inversePermutation(int arr[], int size)
{
int i ,j;
// Loop to select Elements one by one
for ( i = 0; i < size; i++)
{
// Loop to print position of element
// where we find an element
for ( j = 0; j < size; j++)
{
// checking the element in
// increasing order
if (arr[j] == i + 1)
{
// print position of element
// where element is in inverse
// permutation
System.out.print( j + 1 + " ");
break;
}
}
}
}
// Driver program to test above function
public static void main (String[] args)
{
int arr[] = {2, 3, 4, 5, 1};
int size = arr.length;
inversePermutation(arr, size);
}
}
// This code is contributed by vt_m
Python 3
# Naive Python3 Program to
# find inverse permutation.
# Function to find inverse permutations
def inversePermutation(arr, size):
# Loop to select Elements one by one
for i in range(0, size):
# Loop to print position of element
# where we find an element
for j in range(0, size):
# checking the element in increasing order
if (arr[j] == i + 1):
# print position of element where
# element is in inverse permutation
print(j + 1, end = " ")
break
# Driver Code
arr = [2, 3, 4, 5, 1]
size = len(arr)
inversePermutation(arr, size)
#This code is contributed by Smitha Dinesh Semwal
C
// Naive C# Program to find inverse permutation.
using System;
class GFG {
// java function to find inverse permutations
static void inversePermutation(int []arr, int size)
{
int i ,j;
// Loop to select Elements one by one
for ( i = 0; i < size; i++)
{
// Loop to print position of element
// where we find an element
for ( j = 0; j < size; j++)
{
// checking the element in
// increasing order
if (arr[j] == i + 1)
{
// print position of element
// where element is in inverse
// permutation
Console.Write( j + 1 + " ");
break;
}
}
}
}
// Driver program to test above function
public static void Main ()
{
int []arr = {2, 3, 4, 5, 1};
int size = arr.Length;
inversePermutation(arr, size);
}
}
// This code is contributed by vt_m
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Naive PHP Program to
// find inverse permutation.
// Function to find
// inverse permutations
function inversePermutation($arr, $size)
{
for ( $i = 0; $i < $size; $i++)
{
// Loop to print position of element
// where we find an element
for ($j = 0; $j < $size; $j++)
{
// checking the element
// in increasing order
if ($arr[$j] == $i + 1)
{
// print position of element
// where element is in
// inverse permutation
echo $j + 1 , " ";
break;
}
}
}
}
// Driver Code
$arr = array(2, 3, 4, 5, 1);
$size = sizeof($arr);
inversePermutation($arr, $size);
// This code is contributed by aj_36
?>
java 描述语言
<script>
// Naive JavaScript program to find inverse permutation.
// JavaScript function to find inverse permutations
function inversePermutation(arr, size)
{
let i ,j;
// Loop to select Elements one by one
for ( i = 0; i < size; i++)
{
// Loop to print position of element
// where we find an element
for ( j = 0; j < size; j++)
{
// checking the element in
// increasing order
if (arr[j] == i + 1)
{
// print position of element
// where element is in inverse
// permutation
document.write( j + 1 + " ");
break;
}
}
}
}
// Driver code
let arr = [2, 3, 4, 5, 1];
let size = arr.length;
inversePermutation(arr, size);
</script>
输出:
5 1 2 3 4
方法 2 : 想法是使用另一个数组来存储索引和元素映射
C++
// Efficient CPP Program to find inverse permutation.
#include <bits/stdc++.h>
using namespace std;
// C++ function to find inverse permutations
void inversePermutation(int arr[], int size) {
// to store element to index mappings
int arr2[size];
// Inserting position at their
// respective element in second array
for (int i = 0; i < size; i++)
arr2[arr[i] - 1] = i + 1;
for (int i = 0; i < size; i++)
cout << arr2[i] << " ";
}
// Driver program to test above function
int main() {
int arr[] = {2, 3, 4, 5, 1};
int size = sizeof(arr) / sizeof(arr[0]);
inversePermutation(arr, size);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Efficient Java Program to find
// inverse permutation.
import java.io.*;
class GFG {
// function to find inverse permutations
static void inversePermutation(int arr[], int size) {
// to store element to index mappings
int arr2[] = new int[size];
// Inserting position at their
// respective element in second array
for (int i = 0; i < size; i++)
arr2[arr[i] - 1] = i + 1;
for (int i = 0; i < size; i++)
System.out.print(arr2[i] + " ");
}
// Driver program to test above function
public static void main(String args[]) {
int arr[] = {2, 3, 4, 5, 1};
int size = arr.length;
inversePermutation(arr, size);
}
}
// This code is contributed by Nikita Tiwari.
Python 3
# Efficient Python 3 Program to find
# inverse permutation.
# function to find inverse permutations
def inversePermutation(arr, size) :
# To store element to index mappings
arr2 = [0] *(size)
# Inserting position at their
# respective element in second array
for i in range(0, size) :
arr2[arr[i] - 1] = i + 1
for i in range(0, size) :
print( arr2[i], end = " ")
# Driver program
arr = [2, 3, 4, 5, 1]
size = len(arr)
inversePermutation(arr, size)
# This code is contributed by Nikita Tiwari.
C
// Efficient C# Program to find
// inverse permutation.
using System;
class GFG {
// function to find inverse permutations
static void inversePermutation(int []arr, int size) {
// to store element to index mappings
int []arr2 = new int[size];
// Inserting position at their
// respective element in second array
for (int i = 0; i < size; i++)
arr2[arr[i] - 1] = i + 1;
for (int i = 0; i < size; i++)
Console.Write(arr2[i] + " ");
}
// Driver program to test above function
public static void Main() {
int []arr = {2, 3, 4, 5, 1};
int size = arr.Length;
inversePermutation(arr, size);
}
}
// This code is contributed by vt_m.
Output : 5 1 2 3 4
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