在给定条件下将一个排列转换为另一个排列所需的最小相邻交换数

给定大小为N的置换P ，其值从1到N。任务是找到所需的最小相邻交换数，以使得对于[1，N]范围内的所有i， P [i]不等于i 。
例子：

Input: P = [1, 4, 3, 5, 2] 
Output: 2 
Explanation: 
Here P = [1, 4, 3, 5, 2] at index 1, 2, 3, 4, 5. As we can see, P[1] = 1 and P[3] = 3. Hence, we need to get rid of this invariant. 
Swap 1: Swap index 1 and index 2 => [4, 1, 3, 5, 2] 
Swap 2: Swap index 2 and index 3 => [4, 3, 1, 5, 2] 
The final array has no i where P[i] = i. Hence, a minimum of 2 swaps is required.
Input: P = [1, 2, 4, 9, 5, 8, 7, 3, 6] 
Output: 3 
Explanation: 
Swap 1: Swap index 1 and index 2 => [2, 1, 4, 9, 5, 8, 7, 3, 6] 
Swap 2: Swap index 5 and index 6 => [2, 1, 4, 9, 8, 5, 7, 3, 6] 
Swap 2: Swap index 7 and index 8 => [2, 1, 4, 9, 8, 5, 3, 7, 6] 
Hence, a minimum of 3 swaps is required. 
 

方法：让我们考虑用X表示P [i] = i的位置，用O表示其他位置。以下是对该问题的三个基本观察：

• 如果排列的任意两个相邻索引处的值的格式为XO ，则我们可以简单地交换这两个索引以获取’OO’。
• 如果该排列的任意两个相邻索引处的值的格式为XX ，则我们可以简单地交换这两个索引以获取’OO’。
• 如果排列的任意两个相邻索引处的值的格式为OX ，则一旦指针到达X处的索引，它的名称就是‘XO’或‘XX’ 。

步骤如下：

1. 从1迭代到N – 1并检查P [i] = i，然后我们简单地交换(P [i]，P [i + 1]) ，否则继续下一个相邻对的处理。
2. 给定问题的极端案例是当i = N时，如果P [i] = i，则我们交换(P [i]，P [i – 1]) 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum
// number of swaps
void solve(vector& P, int n)
{
 
    // New array to convert
    // to 1-based indexing
    vector arr;
 
    arr.push_back(0);
 
    for (auto x : P)
        arr.push_back(x);
 
    // Keeps count of swaps
    int cnt = 0;
 
    for (int i = 1; i < n; i++) {
 
        // Check if it is an 'X' position
        if (arr[i] == i) {
            swap(arr[i], arr[i + 1]);
            cnt++;
        }
    }
 
    // Corner Case
    if (arr[n] == n) {
 
        swap(arr[n - 1], arr[n]);
        cnt++;
    }
 
    // Print the minimum swaps
    cout << cnt << endl;
}
 
// Driver Code
signed main()
{
    // Given Number N
    int N = 9;
 
    // Given Permutation of N numbers
    vector P = { 1, 2, 4, 9, 5,
                      8, 7, 3, 6 };
 
    // Function Call
    solve(P, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find the minimum
// number of swaps
static void solve(int P[], int n)
{
 
    // New array to convert
    // to 1-based indexing
    int arr[] = new int[n + 1];
 
    arr[0] = 0;
 
    for(int i = 0; i < n; i++)
       arr[i + 1] = P[i];
 
    // Keeps count of swaps
    int cnt = 0;
 
    for(int i = 1; i < n; i++)
    {
        
       // Check if it is an 'X' position
       if (arr[i] == i)
       {
           int t = arr[i + 1];
           arr[i + 1] = arr[i];
           arr[i] = t;
           cnt++;
       }
    }
 
    // Corner Case
    if (arr[n] == n)
    {
         
        // Swap
        int t = arr[n - 1];
        arr[n - 1] = arr[n];
        arr[n] = t;
        cnt++;
    }
 
    // Print the minimum swaps
    System.out.println(cnt);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Number N
    int N = 9;
 
    // Given Permutation of N numbers
    int P[] = new int[]{ 1, 2, 4, 9, 5,
                         8, 7, 3, 6 };
 
    // Function Call
    solve(P, N);
}
}
 
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
 
# Function to find the minimum
# number of swaps
def solve(P, n):
 
    # New array to convert
    # to 1-based indexing
    arr = []
 
    arr.append(0)
 
    for x in P:
        arr.append(x)
 
    # Keeps count of swaps
    cnt = 0
 
    for i in range(1, n):
 
        # Check if it is an 'X' position
        if (arr[i] == i):
            arr[i], arr[i + 1] = arr[i + 1], arr[i]
            cnt += 1
 
    # Corner Case
    if (arr[n] == n):
        arr[n - 1], arr[n] = arr[n] , arr[n - 1]
        cnt += 1
 
    # Print the minimum swaps
    print(cnt)
 
# Driver Code
 
# Given number N
N = 9
 
# Given permutation of N numbers
P = [ 1, 2, 4, 9, 5,
      8, 7, 3, 6 ]
 
# Function call
solve(P, N)
 
# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the minimum
// number of swaps
static void solve(int []P, int n)
{
     
    // New array to convert
    // to 1-based indexing
    int []arr = new int[n + 1];
 
    arr[0] = 0;
 
    for(int i = 0; i < n; i++)
        arr[i + 1] = P[i];
 
    // Keeps count of swaps
    int cnt = 0;
 
    for(int i = 1; i < n; i++)
    {
         
        // Check if it is an 'X' position
        if (arr[i] == i)
        {
            int t = arr[i + 1];
            arr[i + 1] = arr[i];
            arr[i] = t;
            cnt++;
        }
    }
 
    // Corner Case
    if (arr[n] == n)
    {
         
        // Swap
        int t = arr[n - 1];
        arr[n - 1] = arr[n];
        arr[n] = t;
        cnt++;
    }
 
    // Print the minimum swaps
    Console.WriteLine(cnt);
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given Number N
    int N = 9;
 
    // Given Permutation of N numbers
    int []P = { 1, 2, 4, 9, 5,
                8, 7, 3, 6 };
 
    // Function Call
    solve(P, N);
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// JavaScript program for the above approach
 
// Function to find the minimum
// number of swaps
function solve(P, n)
{
 
    // New array to convert
    // to 1-based indexing
    let arr = Array.from({length: n+1}, (_, i) => 0);
 
    arr[0] = 0;
 
    for(let i = 0; i < n; i++)
       arr[i + 1] = P[i];
 
    // Keeps count of swaps
    let cnt = 0;
 
    for(let i = 1; i < n; i++)
    {
        
       // Check if it is an 'X' position
       if (arr[i] == i)
       {
           let t = arr[i + 1];
           arr[i + 1] = arr[i];
           arr[i] = t;
           cnt++;
       }
    }
 
    // Corner Case
    if (arr[n] == n)
    {
         
        // Swap
        let t = arr[n - 1];
        arr[n - 1] = arr[n];
        arr[n] = t;
        cnt++;
    }
 
    // Prlet the minimum swaps
    document.write(cnt);
}
 
// Driver Code
 
    // Given Number N
    let N = 9;
 
    // Given Permutation of N numbers
    let P = [ 1, 2, 4, 9, 5,
                         8, 7, 3, 6 ];
 
    // Function Call
    solve(P, N);
 

输出：

3

时间复杂度： O(N)

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