用于生成排列的堆算法

堆算法用于生成n个对象的所有排列。想法是通过选择一对要互换的元素来生成前一个排列的每个排列，而不会干扰其他n-2个元素。
以下是生成n个给定数字的所有排列的说明。
例子：

Input: 1 2 3
Output: 1 2 3
        2 1 3
        3 1 2
        1 3 2
        2 3 1
        3 2 1

算法：

1. 该算法生成(n-1)！前n-1个元素的排列，最后一个元素与每个元素相邻。这将生成以最后一个元素结尾的所有排列。
2. 如果n为奇数，则交换第一个和最后一个元素，如果n为偶数，则交换第i^个元素(i是从0开始的计数器)和最后一个元素，并重复上述算法，直到i小于n。
3. 在每次迭代中，算法都会生成以当前最后一个元素结尾的所有排列。

执行：

C++

// C++ program to print all permutations using
// Heap's algorithm
#include 
using namespace std;
 
// Prints the array
void printArr(int a[], int n)
{
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
    printf("\n");
}
 
// Generating permutation using Heap Algorithm
void heapPermutation(int a[], int size, int n)
{
    // if size becomes 1 then prints the obtained
    // permutation
    if (size == 1) {
        printArr(a, n);
        return;
    }
 
    for (int i = 0; i < size; i++) {
        heapPermutation(a, size - 1, n);
 
        // if size is odd, swap 0th i.e (first) and
        // (size-1)th i.e (last) element
        if (size % 2 == 1)
            swap(a[0], a[size - 1]);
 
        // If size is even, swap ith and
        // (size-1)th i.e (last) element
        else
            swap(a[i], a[size - 1]);
    }
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 3 };
    int n = sizeof a / sizeof a[0];
    heapPermutation(a, n, n);
    return 0;
}

Java

// Java program to print all permutations using
// Heap's algorithm
class HeapAlgo {
    // Prints the array
    void printArr(int a[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(a[i] + " ");
        System.out.println();
    }
 
    // Generating permutation using Heap Algorithm
    void heapPermutation(int a[], int size, int n)
    {
        // if size becomes 1 then prints the obtained
        // permutation
        if (size == 1)
            printArr(a, n);
 
        for (int i = 0; i < size; i++) {
            heapPermutation(a, size - 1, n);
 
            // if size is odd, swap 0th i.e (first) and
            // (size-1)th i.e (last) element
            if (size % 2 == 1) {
                int temp = a[0];
                a[0] = a[size - 1];
                a[size - 1] = temp;
            }
 
            // If size is even, swap ith
            // and (size-1)th i.e last element
            else {
                int temp = a[i];
                a[i] = a[size - 1];
                a[size - 1] = temp;
            }
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        HeapAlgo obj = new HeapAlgo();
        int a[] = { 1, 2, 3 };
        obj.heapPermutation(a, a.length, a.length);
    }
}
 
// This code has been contributed by Amit Khandelwal.

Python3

# Python program to print all permutations using
# Heap's algorithm
 
# Generating permutation using Heap Algorithm
def heapPermutation(a, size):
 
    # if size becomes 1 then prints the obtained
    # permutation
    if size == 1:
        print(a)
        return
 
    for i in range(size):
        heapPermutation(a, size-1)
 
        # if size is odd, swap 0th i.e (first)
        # and (size-1)th i.e (last) element
        # else If size is even, swap ith
        # and (size-1)th i.e (last) element
        if size & 1:
            a[0], a[size-1] = a[size-1], a[0]
        else:
            a[i], a[size-1] = a[size-1], a[i]
 
 
# Driver code
a = [1, 2, 3]
n = len(a)
heapPermutation(a, n)
 
# This code is contributed by ankush_953
# This code was cleaned up to by more pythonic by glubs9

C#

// C# program to print all permutations using
// Heap's algorithm
using System;
 
public class GFG {
    // Prints the array
    static void printArr(int[] a, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(a[i] + " ");
        Console.WriteLine();
    }
 
    // Generating permutation using Heap Algorithm
    static void heapPermutation(int[] a, int size, int n)
    {
        // if size becomes 1 then prints the obtained
        // permutation
        if (size == 1)
            printArr(a, n);
 
        for (int i = 0; i < size; i++) {
            heapPermutation(a, size - 1, n);
 
            // if size is odd, swap 0th i.e (first) and
            // (size-1)th i.e (last) element
            if (size % 2 == 1) {
                int temp = a[0];
                a[0] = a[size - 1];
                a[size - 1] = temp;
            }
 
            // If size is even, swap ith and
            // (size-1)th i.e (last) element
            else {
                int temp = a[i];
                a[i] = a[size - 1];
                a[size - 1] = temp;
            }
        }
    }
 
    // Driver code
    public static void Main()
    {
 
        int[] a = { 1, 2, 3 };
        heapPermutation(a, a.Length, a.Length);
    }
}
 
/* This Java code is contributed by 29AjayKumar*/

输出：

1 2 3
2 1 3
3 1 2
1 3 2
2 3 1
3 2 1