算法题：如何计算排列系数？代码实现

2021年3月23日14:46:57 发表评论 1,003 次浏览

本文概述

排列是指将给定集合中的所有成员排列成一个序列的过程。一个n个元素的集合上的排列数由n!给出,"!"表示的阶乘。

用P(n, k)表示的排列系数用来表示从一个n个元素的集合中得到一个有k个元素的有序子集的方法的个数。

从数学上讲, 它为：

珀姆

图片来源：维基

例子 ：

P(10, 2) = 90
P(10, 3) = 720
P(10, 0) = 1
P(10, 1) = 10

也可以使用以下递归公式来递归计算系数：

P(n, k) = P(n-1, k) + k* P(n-1, k-1)

如果仔细观察, 我们可以分析问题具有重叠的子结构, 因此可以在此处应用动态编程。下面是实现相同想法的程序。

C

// A Dynamic Programming based 
// solution that uses table P[][]
// to calculate the Permutation
// Coefficient
#include<bits/stdc++.h>
  
// Returns value of Permutation
// Coefficient P(n, k)
int permutationCoeff( int n, int k)
{
     int P[n + 1][k + 1];
  
     // Calculate value of Permutation 
     // Coefficient in bottom up manner
     for ( int i = 0; i <= n; i++)
     {
         for ( int j = 0; j <= std::min(i, k); j++)
         {
             // Base Cases
             if (j == 0)
                 P[i][j] = 1;
  
             // Calculate value using
             // previosly stored values
             else
                 P[i][j] = P[i - 1][j] + 
                           (j * P[i - 1][j - 1]);
  
             // This step is important
             // as P(i, j)=0 for j>i
             P[i][j + 1] = 0;
         }
     }
     return P[n][k];
}
  
// Driver Code
int main()
{
     int n = 10, k = 2;
     printf ( "Value of P(%d, %d) is %d " , n, k, permutationCoeff(n, k));
     return 0;
}

Java

// Java code for Dynamic Programming based 
// solution that uses table P[][] to 
// calculate the Permutation Coefficient
import java.io.*;
import java.math.*;
  
class GFG 
{
      
     // Returns value of Permutation 
     // Coefficient P(n, k)
     static int permutationCoeff( int n, int k)
     {
         int P[][] = new int [n + 2 ][k + 2 ];
      
         // Calculate value of Permutation 
         // Coefficient in bottom up manner
         for ( int i = 0 ; i <= n; i++)
         {
             for ( int j = 0 ; 
                  j <= Math.min(i, k); 
                  j++)
             {
                 // Base Cases
                 if (j == 0 )
                     P[i][j] = 1 ;
      
                 // Calculate value using previosly 
                 // stored values
                 else
                     P[i][j] = P[i - 1 ][j] + 
                                (j * P[i - 1 ][j - 1 ]);
      
                 // This step is important 
                 // as P(i, j)=0 for j>i
                 P[i][j + 1 ] = 0 ;
             }
         }
         return P[n][k];
     }
      
     // Driver Code
     public static void main(String args[])
     {
         int n = 10 , k = 2 ;
         System.out.println( "Value of P( " + n + ", " + k + ")" + 
                            " is " + permutationCoeff(n, k) );
     }
}
  
// This code is contributed by Nikita Tiwari.

Python3

# A Dynamic Programming based
# solution that uses
# table P[][] to calculate the
# Permutation Coefficient
  
# Returns value of Permutation
# Coefficient P(n, k)
def permutationCoeff(n, k):
  
     P = [[ 0 for i in range (k + 1 )] 
             for j in range (n + 1 )]
  
     # Calculate value of Permutation
     # Coefficient in
     # bottom up manner
     for i in range (n + 1 ):
         for j in range ( min (i, k) + 1 ):
  
             # Base cases
             if (j = = 0 ):
                 P[i][j] = 1
  
             # Calculate value using 
             # previously stored values
             else :
                 P[i][j] = P[i - 1 ][j] + (
                            j * P[i - 1 ][j - 1 ])
  
             # This step is important 
             # as P(i, j) = 0 for j>i
             if (j < k):
                 P[i][j + 1 ] = 0
     return P[n][k]
  
# Driver Code
n = 10
k = 2
print ( "Value fo P(" , n, ", " , k, ") is " , permutationCoeff(n, k), sep = "")
  
# This code is contributed by Soumen Ghosh.

C#

// C# code for Dynamic Programming based 
// solution that uses table P[][] to 
// calculate the Permutation Coefficient
using System;
  
class GFG 
{
      
     // Returns value of Permutation 
     // Coefficient P(n, k)
     static int permutationCoeff( int n, int k)
     {
         int [, ]P = new int [n + 2, k + 2];
      
         // Calculate value of Permutation 
         // Coefficient in bottom up manner
         for ( int i = 0; i <= n; i++)
         {
             for ( int j = 0; 
                 j <= Math.Min(i, k); 
                 j++)
             {
                 // Base Cases
                 if (j == 0)
                     P[i, j] = 1;
      
                 // Calculate value using previosly 
                 // stored values
                 else
                     P[i, j] = P[i - 1, j] + 
                             (j * P[i - 1, j - 1]);
      
                 // This step is important 
                 // as P(i, j)=0 for j>i
                 P[i, j + 1] = 0;
             }
         }
         return P[n, k];
     }
      
     // Driver Code
     public static void Main()
     {
         int n = 10, k = 2;
         Console.WriteLine( "Value of P( " + n + 
                         ", " + k + ")" + " is " + 
                         permutationCoeff(n, k) );
     }
}
  
// This code is contributed by anuj_67..

的PHP

<?php
// A Dynamic Programming based 
// solution that uses table P[][]
// to calculate the Permutation
// Coefficient
  
// Returns value of Permutation
// Coefficient P(n, k)
function permutationCoeff( $n , $k )
{
     $P = array ( array ());
  
     // Calculate value of Permutation 
     // Coefficient in bottom up manner
     for ( $i = 0; $i <= $n ; $i ++)
     {
         for ( $j = 0; $j <= min( $i , $k ); $j ++)
         {
              
             // Base Cases
             if ( $j == 0)
                 $P [ $i ][ $j ] = 1;
  
             // Calculate value using
             // previosly stored values
             else
                 $P [ $i ][ $j ] = $P [ $i - 1][ $j ] + 
                             ( $j * $P [ $i - 1][ $j - 1]);
  
             // This step is important
             // as P(i, j)=0 for j>i
             $P [ $i ][ $j + 1] = 0;
         }
     }
     return $P [ $n ][ $k ];
}
  
     // Driver Code
     $n = 10; $k = 2;
     echo "Value of P(" , $n , " , " , $k , ") is " , permutationCoeff( $n , $k );
  
// This code is contributed by anuj_67.
?>

输出：

Value of P(10, 2) is 90

在这里我们可以看到时间复杂度为O(n * k), 空间复杂度为O(n * k), 因为程序使用辅助矩阵存储结果。

我们可以在O(n)时间内完成吗？

让我们假设我们维护一个1D数组来计算最多n个阶乘。我们可以使用计算的阶乘值并应用公式P(n, k)= n！ /(n-k)!。以下是说明相同概念的程序。

C ++

// A O(n) solution that uses 
// table fact[] to calculate 
// the Permutation Coefficient
#include<bits/stdc++.h>
using namespace std;
  
// Returns value of Permutation
// Coefficient P(n, k)
int permutationCoeff( int n, int k)
{
     int fact[n + 1];
  
     // Base case
     fact[0] = 1;
  
     // Calculate value 
     // factorials up to n
     for ( int i = 1; i <= n; i++)
        fact[i] = i * fact[i - 1];
  
     // P(n, k) = n! / (n - k)!
     return fact[n] / fact[n - k];
}
  
// Driver Code
int main()
{
     int n = 10, k = 2;
      
     cout << "Value of P(" << n << ", "
          << k << ") is "  
          << permutationCoeff(n, k);
  
     return 0;
}
  
// This code is contributed by shubhamsingh10

C

// A O(n) solution that uses 
// table fact[] to calculate 
// the Permutation Coefficient
#include<bits/stdc++.h>
  
// Returns value of Permutation
// Coefficient P(n, k)
int permutationCoeff( int n, int k)
{
     int fact[n + 1];
  
     // base case
     fact[0] = 1;
  
     // Calculate value 
     // factorials up to n
     for ( int i = 1; i <= n; i++)
         fact[i] = i * fact[i - 1];
  
     // P(n, k) = n! / (n - k)!
     return fact[n] / fact[n - k];
}
  
// Driver Code
int main()
{
     int n = 10, k = 2;
     printf ( "Value of P(%d, %d) is %d " , n, k, permutationCoeff(n, k) );
     return 0;
}

Java

// A O(n) solution that uses 
// table fact[] to calculate 
// the Permutation Coefficient
import java .io.*;
  
public class GFG {
      
     // Returns value of Permutation
     // Coefficient P(n, k)
     static int permutationCoeff( int n, int k)
     {
         int []fact = new int [n+ 1 ];
      
         // base case
         fact[ 0 ] = 1 ;
      
         // Calculate value 
         // factorials up to n
         for ( int i = 1 ; i <= n; i++)
             fact[i] = i * fact[i - 1 ];
      
         // P(n, k) = n! / (n - k)!
         return fact[n] / fact[n - k];
     }
      
     // Driver Code
     static public void main (String[] args)
     {
         int n = 10 , k = 2 ;
         System.out.println( "Value of"
         + " P( " + n + ", " + k + ") is "
         + permutationCoeff(n, k) );
     }
}
  
// This code is contributed by anuj_67.

Python3

# A O(n) solution that uses 
# table fact[] to calculate 
# the Permutation Coefficient
  
# Returns value of Permutation
# Coefficient P(n, k)
def permutationCoeff(n, k):
     fact = [ 0 for i in range (n + 1 )]
  
     # base case
     fact[ 0 ] = 1
  
     # Calculate value 
     # factorials up to n
     for i in range ( 1 , n + 1 ):
         fact[i] = i * fact[i - 1 ]
  
     # P(n, k) = n!/(n-k)!
     return int (fact[n] / fact[n - k])
  
# Driver Code
n = 10
k = 2
print ( "Value of P(" , n, ", " , k, ") is " , permutationCoeff(n, k), sep = "")
  
# This code is contributed
# by Soumen Ghosh

C#

// A O(n) solution that uses 
// table fact[] to calculate 
// the Permutation Coefficient
using System;
  
public class GFG {
      
     // Returns value of Permutation
     // Coefficient P(n, k)
     static int permutationCoeff( int n, int k)
     {
         int []fact = new int [n+1];
      
         // base case
         fact[0] = 1;
      
         // Calculate value 
         // factorials up to n
         for ( int i = 1; i <= n; i++)
             fact[i] = i * fact[i - 1];
      
         // P(n, k) = n! / (n - k)!
         return fact[n] / fact[n - k];
     }
      
     // Driver Code
     static public void Main ()
     {
         int n = 10, k = 2;
         Console.WriteLine( "Value of"
         + " P( " + n + ", " + k + ") is "
          + permutationCoeff(n, k) );
     }
}
  
// This code is contributed by anuj_67.

的PHP

<?php
// A O(n) Solution that 
// uses table fact[] to 
// calculate the Permutation 
// Coefficient
  
// Returns value of Permutation
// Coefficient P(n, k)
function permutationCoeff( $n , $k )
{
     $fact = array ();
  
     // base case
     $fact [0] = 1;
  
     // Calculate value 
     // factorials up to n
     for ( $i = 1; $i <= $n ; $i ++)
         $fact [ $i ] = $i * $fact [ $i - 1];
  
     // P(n, k)= n!/(n-k)!
     return $fact [ $n ] / $fact [ $n - $k ];
}
  
     // Driver Code
     $n = 10;
     $k = 2;
     echo "Value of P(" , $n , " " , $k , ") is " , permutationCoeff( $n , $k ) ;
              
// This code is contributed by anuj_67.             
?>

输出：

Value of P(10, 2) is 90

O(n)时间和O(1)额外空间解决方案

C

// A O(n) time and O(1) extra
// space solution to calculate
// the Permutation Coefficient
#include <iostream>
using namespace std;
  
int PermutationCoeff( int n, int k)
{
     int P = 1;
  
     // Compute n*(n-1)*(n-2)....(n-k+1)
     for ( int i = 0; i < k; i++)
         P *= (n-i) ;
  
     return P;
}
  
// Driver Code
int main()
{
     int n = 10, k = 2;
     cout << "Value of P(" << n << ", " << k
          << ") is " << PermutationCoeff(n, k);
  
     return 0;
}

Java

// A O(n) time and O(1) extra 
// space solution to calculate
// the Permutation Coefficient
import java.io.*;
  
class GFG 
{
     static int PermutationCoeff( int n, int k)
     {
         int Fn = 1 , Fk = 1 ;
      
         // Compute n! and (n-k)!
         for ( int i = 1 ; i <= n; i++)
         {
             Fn *= i;
             if (i == n - k)
             Fk = Fn;
         }
         int coeff = Fn / Fk;
         return coeff;
     }
      
     // Driver Code 
     public static void main(String args[])
     {
         int n = 10 , k = 2 ;
         System.out.println( "Value of P( " + n + ", " + 
                                          k + ") is " + 
                             PermutationCoeff(n, k) );
     }
}
      
// This code is contributed by Nikita Tiwari.

Python3

# A O(n) time and O(1) extra 
# space solution to calculate
# the Permutation Coefficient
  
def PermutationCoeff(n, k):
     Fn = 1
  
     # Compute n! and (n-k)!
     for i in range ( 1 , n + 1 ):
         Fn * = i
         if (i = = n - k):
             Fk = Fn
  
     coeff = Fn / / Fk
     return coeff
  
# Driver Code
n = 10
k = 2
print ( "Value of P(" , n, ", " , k, ") is " , PermutationCoeff(n, k), sep = "")
  
# This code is contributed
# by Soumen Ghosh.

C#

// A O(n) time and O(1) extra 
// space solution to calculate
// the Permutation Coefficient
using System;
  
class GFG {
      
     static int PermutationCoeff( int n, int k)
     {
         int Fn = 1, Fk = 1;
      
         // Compute n! and (n-k)!
         for ( int i = 1; i <= n; i++)
         {
             Fn *= i;
             if (i == n - k)
                 Fk = Fn;
         }
         int coeff = Fn / Fk;
          
         return coeff;
     }
      
     // Driver Code 
     public static void Main()
     {
         int n = 10, k = 2;
         Console.WriteLine( "Value of P( "
                    + n + ", " + k + ") is "
               + PermutationCoeff(n, k) );
     }
}
      
// This code is contributed by anuj_67.

的PHP

<?php
// A O(n) time and O(1) extra 
// space PHP solution to calculate 
// the Permutation Coefficient
  
function PermutationCoeff( $n , $k )
{
     $Fn = 1; $Fk ;
  
     // Compute n! and (n-k)!
     for ( $i = 1; $i <= $n ; $i ++)
     {
         $Fn *= $i ;
         if ( $i == $n - $k )
         $Fk = $Fn ;
     }
     $coeff = $Fn / $Fk ;
     return $coeff ;
}
  
// Driver Code
$n = 10; $k = 2;
echo "Value of P(" , $n , ", " , $k , ")
         is " , PermutationCoeff( $n , $k );
  
// This code is contributed by anuj_67. 
?>

输出：

Value of P(10, 2) is 90

感谢Shiva Kumar提出此解决方案。

本文作者：阿舒托什·库马尔(Ashutosh Kumar)。如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请发表评论。

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