关于c#:生成集合的排列(最有效)

Generating permutations of a set (most efficiently)

我想生成集合(集合)的所有排列,如下所示:

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Collection: 1, 2, 3
Permutations: {1, 2, 3}
              {1, 3, 2}
              {2, 1, 3}
              {2, 3, 1}
              {3, 1, 2}
              {3, 2, 1}

一般来说,这不是一个"如何"的问题,而是关于如何最有效的问题。另外,我不希望生成所有排列并返回它们,但一次只生成一个排列,并且仅在必要时继续(很像迭代器——我也尝试过,但结果效率较低)。

我已经测试了许多算法和方法,并提出了这一代码,这是我尝试过的最有效的代码:

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public static bool NextPermutation<T>(T[] elements) where T : IComparable<T>
{
    // More efficient to have a variable instead of accessing a property
    var count = elements.Length;

    // Indicates whether this is the last lexicographic permutation
    var done = true;

    // Go through the array from last to first
    for (var i = count - 1; i > 0; i--)
    {
        var curr = elements[i];

        // Check if the current element is less than the one before it
        if (curr.CompareTo(elements[i - 1]) < 0)
        {
            continue;
        }

        // An element bigger than the one before it has been found,
        // so this isn't the last lexicographic permutation.
        done = false;

        // Save the previous (bigger) element in a variable for more efficiency.
        var prev = elements[i - 1];

        // Have a variable to hold the index of the element to swap
        // with the previous element (the to-swap element would be
        // the smallest element that comes after the previous element
        // and is bigger than the previous element), initializing it
        // as the current index of the current item (curr).
        var currIndex = i;

        // Go through the array from the element after the current one to last
        for (var j = i + 1; j < count; j++)
        {
            // Save into variable for more efficiency
            var tmp = elements[j];

            // Check if tmp suits the"next swap" conditions:
            // Smallest, but bigger than the"prev" element
            if (tmp.CompareTo(curr) < 0 && tmp.CompareTo(prev) > 0)
            {
                curr = tmp;
                currIndex = j;
            }
        }

        // Swap the"prev" with the new"curr" (the swap-with element)
        elements[currIndex] = prev;
        elements[i - 1] = curr;

        // Reverse the order of the tail, in order to reset it's lexicographic order
        for (var j = count - 1; j > i; j--, i++)
        {
            var tmp = elements[j];
            elements[j] = elements[i];
            elements[i] = tmp;
        }

        // Break since we have got the next permutation
        // The reason to have all the logic inside the loop is
        // to prevent the need of an extra variable indicating"i" when
        // the next needed swap is found (moving"i" outside the loop is a
        // bad practice, and isn't very readable, so I preferred not doing
        // that as well).
        break;
    }

    // Return whether this has been the last lexicographic permutation.
    return done;
}

它的用法是发送一个元素数组,并返回一个布尔值,指示这是否是最后一个词典编纂排列,以及将数组更改为下一个排列。

使用实例:

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var arr = new[] {1, 2, 3};

PrintArray(arr);

while (!NextPermutation(arr))
{
    PrintArray(arr);
}

问题是我对代码的速度不满意。

遍历大小为11的数组的所有排列大约需要4秒钟。虽然可以认为这是令人印象深刻的,因为一组11号的可能排列的数量是11!,接近4000万。

从逻辑上讲,对于一个12大小的数组,它将花费大约12倍的时间,因为12!11! * 12,而对于13大小的数组,它将花费大约13倍于12大小的时间,依此类推。

因此,您可以很容易地理解,对于大小为12或更多的数组,执行所有排列确实需要很长的时间。

我有一种强烈的预感,我可以以某种方式将时间缩短很多(不用切换到C语言以外的语言),因为编译器优化确实非常好地优化了,我怀疑我可以在汇编中手动优化。

有没有人知道其他更快的方法?你知道如何使当前算法更快吗?

请注意,我不想使用外部库或服务来实现这一点——我想拥有代码本身,我希望它尽可能地高效。


这可能是你要找的。

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    private static bool NextPermutation(int[] numList)
    {
        /*
         Knuths
         1. Find the largest index j such that a[j] < a[j + 1]. If no such index exists, the permutation is the last permutation.
         2. Find the largest index l such that a[j] < a[l]. Since j + 1 is such an index, l is well defined and satisfies j < l.
         3. Swap a[j] with a[l].
         4. Reverse the sequence from a[j + 1] up to and including the final element a[n].

         */

        var largestIndex = -1;
        for (var i = numList.Length - 2; i >= 0; i--)
        {
            if (numList[i] < numList[i + 1]) {
                largestIndex = i;
                break;
            }
        }

        if (largestIndex < 0) return false;

        var largestIndex2 = -1;
        for (var i = numList.Length - 1 ; i >= 0; i--) {
            if (numList[largestIndex] < numList[i]) {
                largestIndex2 = i;
                break;
            }
        }

        var tmp = numList[largestIndex];
        numList[largestIndex] = numList[largestIndex2];
        numList[largestIndex2] = tmp;

        for (int i = largestIndex + 1, j = numList.Length - 1; i < j; i++, j--) {
            tmp = numList[i];
            numList[i] = numList[j];
            numList[j] = tmp;
        }

        return true;
    }


Update 2018-05-28:

  • A new multithreaded version (lot faster) is available below as another answer.
  • Also an article about permutation: Permutations: Fast implementations and a new indexing algorithm allowing multithreading

有点晚了…

根据最近的测试(更新日期:2018-05-22)

  • 最快的是我的,但不是按字典顺序
  • 对于最快的词典编纂顺序,sani singh huttunen解决方案似乎是可行的方法。

10项(10!)在我的机器上释放(毫秒):

  • Ouellet:29
  • SimpleVar:95
  • 埃雷兹·罗宾逊:156
  • 萨尼·辛格·胡特宁:37岁
  • 彭阳:45047

13项(13!)在我的机器上释放(秒):

  • 欧勒特:48.437
  • 单工:159.869
  • 埃雷兹·罗宾逊:327.781
  • 萨尼·辛格·胡特宁:64.839

我的解决方案的优点:

  • 堆算法(每排列一次交换)
  • 没有乘法(就像在网络上看到的一些实现一样)
  • 内联交换
  • 通用的
  • 无不安全代码
  • 到位(内存使用率非常低)
  • 无模(仅第一位比较)

我对堆算法的实现:

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using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Runtime.CompilerServices;

namespace WpfPermutations
{
    /// <summary>
    /// EO: 2016-04-14
    /// Generator of all permutations of an array of anything.
    /// Base on Heap's Algorithm. See: https://en.wikipedia.org/wiki/Heap%27s_algorithm#cite_note-3
    /// </summary>
    public static class Permutations
    {
        /// <summary>
        /// Heap's algorithm to find all pmermutations. Non recursive, more efficient.
        /// </summary>
        /// <param name="items">Items to permute in each possible ways</param>
        /// <param name="funcExecuteAndTellIfShouldStop"></param>
        /// <returns>Return true if cancelled</returns>
        public static bool ForAllPermutation<T>(T[] items, Func<T[], bool> funcExecuteAndTellIfShouldStop)
        {
            int countOfItem = items.Length;

            if (countOfItem <= 1)
            {
                return funcExecuteAndTellIfShouldStop(items);
            }

            var indexes = new int[countOfItem];
            for (int i = 0; i < countOfItem; i++)
            {
                indexes[i] = 0;
            }

            if (funcExecuteAndTellIfShouldStop(items))
            {
                return true;
            }

            for (int i = 1; i < countOfItem;)
            {
                if (indexes[i] < i)
                { // On the web there is an implementation with a multiplication which should be less efficient.
                    if ((i & 1) == 1) // if (i % 2 == 1)  ... more efficient ??? At least the same.
                    {
                        Swap(ref items[i], ref items[indexes[i]]);
                    }
                    else
                    {
                        Swap(ref items[i], ref items[0]);
                    }

                    if (funcExecuteAndTellIfShouldStop(items))
                    {
                        return true;
                    }

                    indexes[i]++;
                    i = 1;
                }
                else
                {
                    indexes[i++] = 0;
                }
            }

            return false;
        }

        /// <summary>
        /// This function is to show a linq way but is far less efficient
        /// From: StackOverflow user: Pengyang : http://stackoverflow.com/questions/756055/listing-all-permutations-of-a-string-integer
        /// </summary>
        /// <typeparam name="T"></typeparam>
        /// <param name="list"></param>
        /// <param name="length"></param>
        /// <returns></returns>
        static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> list, int length)
        {
            if (length == 1) return list.Select(t => new T[] { t });

            return GetPermutations(list, length - 1)
                .SelectMany(t => list.Where(e => !t.Contains(e)),
                    (t1, t2) => t1.Concat(new T[] { t2 }));
        }

        /// <summary>
        /// Swap 2 elements of same type
        /// </summary>
        /// <typeparam name="T"></typeparam>
        /// <param name="a"></param>
        /// <param name="b"></param>
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        static void Swap<T>(ref T a, ref T b)
        {
            T temp = a;
            a = b;
            b = temp;
        }

        /// <summary>
        /// Func to show how to call. It does a little test for an array of 4 items.
        /// </summary>
        public static void Test()
        {
            ForAllPermutation("123".ToCharArray(), (vals) =>
            {
                Console.WriteLine(String.Join("", vals));
                return false;
            });

            int[] values = new int[] { 0, 1, 2, 4 };

            Console.WriteLine("Ouellet heap's algorithm implementation");
            ForAllPermutation(values, (vals) =>
            {
                Console.WriteLine(String.Join("", vals));
                return false;
            });

            Console.WriteLine("Linq algorithm");
            foreach (var v in GetPermutations(values, values.Length))
            {
                Console.WriteLine(String.Join("", v));
            }

            // Performance Heap's against Linq version : huge differences
            int count = 0;

            values = new int[10];
            for (int n = 0; n < values.Length; n++)
            {
                values[n] = n;
            }

            Stopwatch stopWatch = new Stopwatch();

            ForAllPermutation(values, (vals) =>
            {
                foreach (var v in vals)
                {
                    count++;
                }
                return false;
            });

            stopWatch.Stop();
            Console.WriteLine($"Ouellet heap's algorithm implementation {count} items in {stopWatch.ElapsedMilliseconds} millisecs");

            count = 0;
            stopWatch.Reset();
            stopWatch.Start();

            foreach (var vals in GetPermutations(values, values.Length))
            {
                foreach (var v in vals)
                {
                    count++;
                }
            }

            stopWatch.Stop();
            Console.WriteLine($"Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs");
        }
    }
}

A这是我的测试代码:

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Task.Run(() =>
            {

                int[] values = new int[12];
                for (int n = 0; n < values.Length; n++)
                {
                    values[n] = n;
                }

                // Eric Ouellet Algorithm
                int count = 0;
                var stopwatch = new Stopwatch();
                stopwatch.Reset();
                stopwatch.Start();
                Permutations.ForAllPermutation(values, (vals) =>
                {
                    foreach (var v in vals)
                    {
                        count++;
                    }
                    return false;
                });
                stopwatch.Stop();
                Console.WriteLine($"This {count} items in {stopwatch.ElapsedMilliseconds} millisecs");

                // Simple Plan Algorithm
                count = 0;
                stopwatch.Reset();
                stopwatch.Start();
                PermutationsSimpleVar permutations2 = new PermutationsSimpleVar();
                permutations2.Permutate(1, values.Length, (int[] vals) =>
                {
                    foreach (var v in vals)
                    {
                        count++;
                    }
                });
                stopwatch.Stop();
                Console.WriteLine($"Simple Plan {count} items in {stopwatch.ElapsedMilliseconds} millisecs");

                // ErezRobinson Algorithm
                count = 0;
                stopwatch.Reset();
                stopwatch.Start();
                foreach(var vals in PermutationsErezRobinson.QuickPerm(values))
                {
                    foreach (var v in vals)
                    {
                        count++;
                    }
                };
                stopwatch.Stop();
                Console.WriteLine($"Erez Robinson {count} items in {stopwatch.ElapsedMilliseconds} millisecs");
            });

使用实例:

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ForAllPermutation("123".ToCharArray(), (vals) =>
    {
        Console.WriteLine(String.Join("", vals));
        return false;
    });

int[] values = new int[] { 0, 1, 2, 4 };
ForAllPermutation(values, (vals) =>
        {
            Console.WriteLine(String.Join("", vals));
            return false;
        });


好吧,如果你能用C语言处理,然后翻译成你选择的语言,你就不能比这快得多,因为时间将由print支配:

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void perm(char* s, int n, int i){
  if (i >= n-1) print(s);
  else {
    perm(s, n, i+1);
    for (int j = i+1; j<n; j++){
      swap(s[i], s[j]);
      perm(s, n, i+1);
      swap(s[i], s[j]);
    }
  }
}

perm("ABC", 3, 0);


我所知道的最快置换算法是快速置换算法。这里是实现,它使用yield-return,因此您可以根据需要一次迭代一次。

代码:

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public static IEnumerable<IEnumerable<T>> QuickPerm<T>(this IEnumerable<T> set)
    {
        int N = set.Count();
        int[] a = new int[N];
        int[] p = new int[N];

        var yieldRet = new T[N];

        List<T> list = new List<T>(set);

        int i, j, tmp; // Upper Index i; Lower Index j

        for (i = 0; i < N; i++)
        {
            // initialize arrays; a[N] can be any type
            a[i] = i + 1; // a[i] value is not revealed and can be arbitrary
            p[i] = 0; // p[i] == i controls iteration and index boundaries for i
        }
        yield return list;
        //display(a, 0, 0);   // remove comment to display array a[]
        i = 1; // setup first swap points to be 1 and 0 respectively (i & j)
        while (i < N)
        {
            if (p[i] < i)
            {
                j = i%2*p[i]; // IF i is odd then j = p[i] otherwise j = 0
                tmp = a[j]; // swap(a[j], a[i])
                a[j] = a[i];
                a[i] = tmp;

                //MAIN!

                for (int x = 0; x < N; x++)
                {
                    yieldRet[x] = list[a[x]-1];
                }
                yield return yieldRet;
                //display(a, j, i); // remove comment to display target array a[]

                // MAIN!

                p[i]++; // increase index"weight" for i by one
                i = 1; // reset index i to 1 (assumed)
            }
            else
            {
                // otherwise p[i] == i
                p[i] = 0; // reset p[i] to zero
                i++; // set new index value for i (increase by one)
            } // if (p[i] < i)
        } // while(i < N)
    }


下面是我最后得到的最快的实现:

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public class Permutations
{
    private readonly Mutex _mutex = new Mutex();

    private Action<int[]> _action;
    private Action<IntPtr> _actionUnsafe;
    private unsafe int* _arr;
    private IntPtr _arrIntPtr;
    private unsafe int* _last;
    private unsafe int* _lastPrev;
    private unsafe int* _lastPrevPrev;

    public int Size { get; private set; }

    public bool IsRunning()
    {
        return this._mutex.SafeWaitHandle.IsClosed;
    }

    public bool Permutate(int start, int count, Action<int[]> action, bool async = false)
    {
        return this.Permutate(start, count, action, null, async);
    }

    public bool Permutate(int start, int count, Action<IntPtr> actionUnsafe, bool async = false)
    {
        return this.Permutate(start, count, null, actionUnsafe, async);
    }

    private unsafe bool Permutate(int start, int count, Action<int[]> action, Action<IntPtr> actionUnsafe, bool async = false)
    {
        if (!this._mutex.WaitOne(0))
        {
            return false;
        }

        var x = (Action)(() =>
                             {
                                 this._actionUnsafe = actionUnsafe;
                                 this._action = action;

                                 this.Size = count;

                                 this._arr = (int*)Marshal.AllocHGlobal(count * sizeof(int));
                                 this._arrIntPtr = new IntPtr(this._arr);

                                 for (var i = 0; i < count - 3; i++)
                                 {
                                     this._arr[i] = start + i;
                                 }

                                 this._last = this._arr + count - 1;
                                 this._lastPrev = this._last - 1;
                                 this._lastPrevPrev = this._lastPrev - 1;

                                 *this._last = count - 1;
                                 *this._lastPrev = count - 2;
                                 *this._lastPrevPrev = count - 3;

                                 this.Permutate(count, this._arr);
                             });

        if (!async)
        {
            x();
        }
        else
        {
            new Thread(() => x()).Start();
        }

        return true;
    }

    private unsafe void Permutate(int size, int* start)
    {
        if (size == 3)
        {
            this.DoAction();
            Swap(this._last, this._lastPrev);
            this.DoAction();
            Swap(this._last, this._lastPrevPrev);
            this.DoAction();
            Swap(this._last, this._lastPrev);
            this.DoAction();
            Swap(this._last, this._lastPrevPrev);
            this.DoAction();
            Swap(this._last, this._lastPrev);
            this.DoAction();

            return;
        }

        var sizeDec = size - 1;
        var startNext = start + 1;
        var usedStarters = 0;

        for (var i = 0; i < sizeDec; i++)
        {
            this.Permutate(sizeDec, startNext);

            usedStarters |= 1 << *start;

            for (var j = startNext; j <= this._last; j++)
            {
                var mask = 1 << *j;

                if ((usedStarters & mask) != mask)
                {
                    Swap(start, j);
                    break;
                }
            }
        }

        this.Permutate(sizeDec, startNext);

        if (size == this.Size)
        {
            this._mutex.ReleaseMutex();
        }
    }

    private unsafe void DoAction()
    {
        if (this._action == null)
        {
            if (this._actionUnsafe != null)
            {
                this._actionUnsafe(this._arrIntPtr);
            }

            return;
        }

        var result = new int[this.Size];

        fixed (int* pt = result)
        {
            var limit = pt + this.Size;
            var resultPtr = pt;
            var arrayPtr = this._arr;

            while (resultPtr < limit)
            {
                *resultPtr = *arrayPtr;
                resultPtr++;
                arrayPtr++;
            }
        }

        this._action(result);
    }

    private static unsafe void Swap(int* a, int* b)
    {
        var tmp = *a;
        *a = *b;
        *b = tmp;
    }
}

使用和测试性能:

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var perms = new Permutations();

var sw1 = Stopwatch.StartNew();

perms.Permutate(0,
                11,
                (Action<int[]>)null); // Comment this line and...
                //PrintArr); // Uncomment this line, to print permutations

sw1.Stop();
Console.WriteLine(sw1.Elapsed);

打印方法:

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private static void PrintArr(int[] arr)
{
    Console.WriteLine(string.Join(",", arr));
}

深入:

我甚至很久没有考虑过这个问题,所以我只能解释我的代码,但总的来说:

  • 排列不是词典-这让我实际上可以在排列之间执行较少的操作。
  • 实现是递归的,当"视图"的大小为3时,它跳过复杂的逻辑,只执行6个交换来获取6个排列(或者子排列,如果愿意的话)。
  • 因为排列不在词典编纂的顺序中,我如何决定将哪个元素带到当前"视图"(子排列)的开头?在当前的子排列递归调用中,我保存了已经用作"启动器"的元素的记录,并简单地线性搜索数组尾部没有使用的元素。
  • 该实现仅用于整数,因此要对元素的一般集合进行排列,只需使用排列类来排列索引,而不是实际的集合。
  • 互斥对象的存在只是为了确保在异步执行时不会出错(请注意,您可以传递一个非安全参数,该参数将反过来获得指向排列数组的指针)。不能更改该数组(指针)中元素的顺序!如果您想,您应该将数组复制到tmp数组,或者只使用安全操作参数,它为您处理这个问题-传递的数组已经是一个副本)。
  • 注:

    我不知道这个实现到底有多好——我很久没接触过它了。单独测试和比较其他实现,如果您有任何反馈,请告诉我!

    享受。


    这里是一个通用的排列查找工具,它将遍历集合的每个排列并调用一个计算函数。如果Evaluation函数返回true(它找到了它正在寻找的答案),排列查找器将停止处理。

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    public class PermutationFinder<T>
    {
        private T[] items;
        private Predicate<T[]> SuccessFunc;
        private bool success = false;
        private int itemsCount;

        public void Evaluate(T[] items, Predicate<T[]> SuccessFunc)
        {
            this.items = items;
            this.SuccessFunc = SuccessFunc;
            this.itemsCount = items.Count();

            Recurse(0);
        }

        private void Recurse(int index)
        {
            T tmp;

            if (index == itemsCount)
                success = SuccessFunc(items);
            else
            {
                for (int i = index; i < itemsCount; i++)
                {
                    tmp = items[index];
                    items[index] = items[i];
                    items[i] = tmp;

                    Recurse(index + 1);

                    if (success)
                        break;

                    tmp = items[index];
                    items[index] = items[i];
                    items[i] = tmp;
                }
            }
        }
    }

    下面是一个简单的实现:

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    class Program
    {
        static void Main(string[] args)
        {
            new Program().Start();
        }

        void Start()
        {
            string[] items = new string[5];
            items[0] ="A";
            items[1] ="B";
            items[2] ="C";
            items[3] ="D";
            items[4] ="E";
            new PermutationFinder<string>().Evaluate(items, Evaluate);
            Console.ReadLine();
        }

        public bool Evaluate(string[] items)
        {
            Console.WriteLine(string.Format("{0},{1},{2},{3},{4}", items[0], items[1], items[2], items[3], items[4]));
            bool someCondition = false;

            if (someCondition)
                return true;  // Tell the permutation finder to stop.

            return false;
        }
    }


    (P)Update 2018-05-28,a new version,the fasstest…(Multi-Threated)(p)(P)字母名称(p)字母名称(P)Need:Sani Singh Huttunen(Fastest Lexico)Solution and my new ouelletlexico3 which support indexing(p)(P)Indexing has 2 main advantages:(p)

    • Allows to get any permutation directly
    • Allows multi-threading(derived from the first advantage)

    (P)Article:互换交易:fast implementations and a new indexing algorithm allowing multithreading(p)(P)On my machine(6 hyperthreed cores:12 threads)Xeon [email protected],tests algorithms running with empty studff to do for 13!Items(Time in Millises):(p)

    • 53071:Ouellet(Implementation of Heap)
    • 65366:Sani Singh Huttunen(Fastest Lexico)
    • 11377:Mix Ouelletlexico3-Sani Singh Huttunen

    (P)a side note:using shares properties/variables between threads for permutation action will strongly impact performance if their usage is modification(read/write).这样做会在三个威胁之间产生"错误共享"。你不会得到预期的表现。我得到了这个行为while testing。我的经验所涉及的问题是,当我努力增加全球变量以计算总的兑换率时。(p)(P)USAGE:(p)字母名称(P)Code:(p)字母名称


    这里是一个基于数组元素交换的复杂度为O(n * n!)1的递归实现。数组是用来自1, 2, ..., n的值初始化的。

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    using System;

    namespace Exercise
    {
    class Permutations
    {
        static void Main(string[] args)
        {
            int setSize = 3;
            FindPermutations(setSize);
        }
        //-----------------------------------------------------------------------------
        /* Method: FindPermutations(n) */
        private static void FindPermutations(int n)
        {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++)
            {
                arr[i] = i + 1;
            }
            int iEnd = arr.Length - 1;
            Permute(arr, iEnd);
        }
        //-----------------------------------------------------------------------------  
        /* Method: Permute(arr) */
        private static void Permute(int[] arr, int iEnd)
        {
            if (iEnd == 0)
            {
                PrintArray(arr);
                return;
            }

            Permute(arr, iEnd - 1);
            for (int i = 0; i < iEnd; i++)
            {
                swap(ref arr[i], ref arr[iEnd]);
                Permute(arr, iEnd - 1);
                swap(ref arr[i], ref arr[iEnd]);
            }
        }
    }
    }

    在每一个递归步骤中,我们将最后一个元素与for循环中局部变量指向的当前元素交换,然后通过增加for循环的局部变量和减少for循环的终止条件来表示交换的唯一性,该循环最初设置为num。数组中元素的误码率,当后者变为零时,我们终止递归。

    以下是助手函数:

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        //-----------------------------------------------------------------------------
        /*
            Method: PrintArray()

        */

        private static void PrintArray(int[] arr, string label ="")
        {
            Console.WriteLine(label);
            Console.Write("{");
            for (int i = 0; i < arr.Length; i++)
            {
                Console.Write(arr[i]);
                if (i < arr.Length - 1)
                {
                    Console.Write(",");
                }
            }
            Console.WriteLine("}");
        }
        //-----------------------------------------------------------------------------

        /*
            Method: swap(ref int a, ref int b)

        */

        private static void swap(ref int a, ref int b)
        {
            int temp = a;
            a = b;
            b = temp;
        }

    1。要打印的n元素有n!排列。


    Steven Skiena的算法设计手册(第二版第14.4章)中有一个关于算法和实现调查的可访问的介绍。

    斯基纳参考D.Knuth。计算机编程艺术,第4卷第2分册:生成所有元组和排列。Addison-Wesley,2005年。


    如果真的有数量级的改进,我会感到惊讶。如果有,那么C需要根本性的改进。此外,对排列进行任何有趣的操作通常要比生成它花费更多的工作。因此,发电成本在总体方案中是微不足道的。

    也就是说,我建议尝试以下几点。您已经尝试了迭代器。但是您是否尝试过让一个函数接受一个闭包作为输入,然后为找到的每个排列调用这个闭包?根据C的内部力学,这可能更快。

    同样,您是否尝试过让一个函数返回一个将迭代特定排列的闭包?

    无论采用哪种方法,都可以进行许多微优化实验。例如,您可以对输入数组进行排序,然后始终知道它的顺序。例如,您可以有一个bools数组,指示该元素是否小于下一个元素,并且您可以只查看该数组而不进行比较。


    我创建了一个比Knuth的算法稍快的算法:

    11 elements:

    mine: 0.39 seconds

    Knuth's: 0.624 seconds

    13 elements:

    mine: 56.615 seconds

    Knuth's: 98.681 seconds

    以下是我在Java中的代码:

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    public static void main(String[] args)
    {
        int n=11;
        int a,b,c,i,tmp;
        int end=(int)Math.floor(n/2);
        int[][] pos = new int[end+1][2];
        int[] perm = new int[n];

        for(i=0;i<n;i++) perm[i]=i;

        while(true)
        {
            //this is where you can use the permutations (perm)
            i=0;
            c=n;

            while(pos[i][1]==c-2 && pos[i][0]==c-1)
            {
                pos[i][0]=0;
                pos[i][1]=0;
                i++;
                c-=2;
            }

            if(i==end) System.exit(0);

            a=(pos[i][0]+1)%c+i;
            b=pos[i][0]+i;

            tmp=perm[b];
            perm[b]=perm[a];
            perm[a]=tmp;

            if(pos[i][0]==c-1)
            {
                pos[i][0]=0;
                pos[i][1]++;
            }
            else
            {
                pos[i][0]++;
            }
        }
    }

    问题是我的算法只适用于奇数个元素。我很快就写了这段代码,所以我很确定有更好的方法来实现我的想法以获得更好的性能,但我现在没有时间来优化它,并在元素数量相等时解决问题。

    它是每个排列的一个交换,它使用一种非常简单的方法来知道要交换哪些元素。

    我在我的博客上写了一个关于代码背后的方法的解释:http://antoinecomeau.blogspot.ca/2015/01/fast-generation-of-all-permutations.html


    作为这个问题的作者,他问的是一个算法:

    [...] generating a single permutation, at a time, and continuing only if necessary

    我建议考虑Steinhaus–Johnson–Trotter算法。

    Steinhaus–Johnson–Trotter算法(维基百科)

    这里解释得很好


    现在是凌晨1点,我在看电视,想到了同样的问题,但用的是字符串值。

    给一个词,找出所有排列。您可以很容易地修改它来处理数组、集合等。

    我花了点时间想出来,但我想到的解决办法是:

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    string word ="abcd";

    List<string> combinations = new List<string>();

    for(int i=0; i<word.Length; i++)
    {
        for (int j = 0; j < word.Length; j++)
        {
            if (i < j)
                combinations.Add(word[i] + word.Substring(j) + word.Substring(0, i) + word.Substring(i + 1, j - (i + 1)));
            else if (i > j)
            {
                if(i== word.Length -1)
                    combinations.Add(word[i] + word.Substring(0, i));
                else
                    combinations.Add(word[i] + word.Substring(0, i) + word.Substring(i + 1));
            }
        }
    }

    这里有与上面相同的代码,但是有一些注释

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    string word ="abcd";

    List<string> combinations = new List<string>();

    //i is the first letter of the new word combination
    for(int i=0; i<word.Length; i++)
    {
        for (int j = 0; j < word.Length; j++)
        {
            //add the first letter of the word, j is past i so we can get all the letters from j to the end
            //then add all the letters from the front to i, then skip over i (since we already added that as the beginning of the word)
            //and get the remaining letters from i+1 to right before j.
            if (i < j)
                combinations.Add(word[i] + word.Substring(j) + word.Substring(0, i) + word.Substring(i + 1, j - (i + 1)));
            else if (i > j)
            {
                //if we're at the very last word no need to get the letters after i
                if(i== word.Length -1)
                    combinations.Add(word[i] + word.Substring(0, i));
                //add i as the first letter of the word, then get all the letters up to i, skip i, and then add all the lettes after i
                else
                    combinations.Add(word[i] + word.Substring(0, i) + word.Substring(i + 1));

            }
        }
    }

    我在罗塞塔密码上找到了这个算法,它真的是我试过的最快的一个。http://rosettacode.org/wiki/permutations c

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    /* Boothroyd method; exactly N! swaps, about as fast as it gets */
    void boothroyd(int *x, int n, int nn, int callback(int *, int))
    {
        int c = 0, i, t;
        while (1) {
            if (n > 2) boothroyd(x, n - 1, nn, callback);
            if (c >= n - 1) return;
     
            i = (n & 1) ? 0 : c;
            c++;
            t = x[n - 1], x[n - 1] = x[i], x[i] = t;
            if (callback) callback(x, nn);
        }
    }
     
    /* entry for Boothroyd method */
    void perm2(int *x, int n, int callback(int*, int))
    {
        if (callback) callback(x, n);
        boothroyd(x, n, n, callback);
    }


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    //+------------------------------------------------------------------+
    //|                                                                  |
    //+------------------------------------------------------------------+
    /**
     * http://marknelson.us/2002/03/01/next-permutation/
     * Rearranges the elements into the lexicographically next greater permutation and returns true.
     * When there are no more greater permutations left, the function eventually returns false.
     */


    // next lexicographical permutation

    template <typename T>
    bool next_permutation(T &arr[], int firstIndex, int lastIndex)
    {
        int i = lastIndex;
        while (i > firstIndex)
        {
            int ii = i--;
            T curr = arr[i];
            if (curr < arr[ii])
            {
                int j = lastIndex;
                while (arr[j] <= curr) j--;
                Swap(arr[i], arr[j]);
                while (ii < lastIndex)
                    Swap(arr[ii++], arr[lastIndex--]);
                return true;
            }
        }
        return false;
    }

    //+------------------------------------------------------------------+
    //|                                                                  |
    //+------------------------------------------------------------------+
    /**
     * Swaps two variables or two array elements.
     * using references/pointers to speed up swapping.
     */

    template<typename T>
    void Swap(T &var1, T &var2)
    {
        T temp;
        temp = var1;
        var1 = var2;
        var2 = temp;
    }

    //+------------------------------------------------------------------+
    //|                                                                  |
    //+------------------------------------------------------------------+
    // driver program to test above function
    #define N 3

    void OnStart()
    {
        int i, x[N];
        for (i = 0; i < N; i++) x[i] = i + 1;

        printf("The %i! possible permutations with %i elements:", N, N);

        do
        {
            printf("%s", ArrayToString(x));

        } while (next_permutation(x, 0, N - 1));

    }

    // Output:
    // The 3! possible permutations with 3 elements:
    //"1,2,3"
    //"1,3,2"
    //"2,1,3"
    //"2,3,1"
    //"3,1,2"
    //"3,2,1"


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    // Permutations are the different ordered arrangements of an n-element
    // array. An n-element array has exactly n! full-length permutations.

    // This iterator object allows to iterate all full length permutations
    // one by one of an array of n distinct elements.

    // The iterator changes the given array in-place.

    // Permutations('ABCD') => ABCD  DBAC  ACDB  DCBA
    //                         BACD  BDAC  CADB  CDBA
    //                         CABD  ADBC  DACB  BDCA
    //                         ACBD  DABC  ADCB  DBCA
    //                         BCAD  BADC  CDAB  CBDA
    //                         CBAD  ABDC  DCAB  BCDA

    // count of permutations = n!

    // Heap's algorithm (Single swap per permutation)
    // http://www.quickperm.org/quickperm.php
    // https://stackoverflow.com/a/36634935/4208440
    // https://en.wikipedia.org/wiki/Heap%27s_algorithm


    // My implementation of Heap's algorithm:

    template<typename T>
    class PermutationsIterator
    {
        int b, e, n;
        int c[32];  /* control array: mixed radix number in rising factorial base.
                    the i-th digit has base i, which means that the digit must be
                    strictly less than i. The first digit is always 0,  the second
                    can be 0 or 1, the third 0, 1 or 2, and so on.
                    ArrayResize isn't strictly necessary, int c[32] would suffice
                    for most practical purposes. Also, it is much faster */


    public:
        PermutationsIterator(T &arr[], int firstIndex, int lastIndex)
        {
            this.b = firstIndex;  // v.begin()
            this.e = lastIndex;   // v.end()
            this.n = e - b + 1;

            ArrayInitialize(c, 0);
        }

        // Rearranges the input array into the next permutation and returns true.
        // When there are no more permutations left, the function returns false.

        bool next(T &arr[])
        {
            // find index to update
            int i = 1;

            // reset all the previous indices that reached the maximum possible values
            while (c[i] == i)
            {
                c[i] = 0;
                ++i;
            }

            // no more permutations left
            if (i == n)
                return false;

            // generate next permutation
            int j = (i & 1) == 1 ? c[i] : 0;    // IF i is odd then j = c[i] otherwise j = 0.
            swap(arr[b + j], arr[b + i]);       // generate a new permutation from previous permutation using a single swap

            // Increment that index
            ++c[i];
            return true;
        }

    };


    如果你只想计算可能的排列数目,你可以避免上面所有的艰苦工作,并使用类似的东西(在C中设计的):

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    public static class ContrivedUtils
    {
        public static Int64 Permutations(char[] array)
        {
            if (null == array || array.Length == 0) return 0;

            Int64 permutations = array.Length;

            for (var pos = permutations; pos > 1; pos--)
                permutations *= pos - 1;

            return permutations;
        }
    }

    你这样称呼它:

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    var permutations = ContrivedUtils.Permutations("1234".ToCharArray());
    // output is: 24
    var permutations = ContrivedUtils.Permutations("123456789".ToCharArray());
    // output is: 362880