根据给定条件将一个排列转换为另一个排列所需的最小相邻交换次数

给定大小为N的排列 P ，其值从1 到 N 。任务是找到所需的最小相邻交换次数，以便对于[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P[i] = i的位置由X表示，其他位置由O 表示。以下是对这个问题的三个基本观察：

• 如果排列的任何两个相邻索引处的值的形式为XO ，我们可以简单地交换 2 个索引以获得 ‘OO’。
• 如果排列的任何两个相邻索引处的值的形式为XX ，我们可以简单地交换 2 个索引以获得 ‘OO’。
• 如果排列的任何两个相邻索引处的值的形式为OX ，则一旦指针到达X 处的索引，它就只是‘XO’或‘XX’ 。

以下是步骤：

1. 从1 到 N – 1迭代并检查是否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过程。
2. 给定问题的极端情况是当i = N 时，如果 P[i] = i，那么我们交换 (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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