30

我需要迭代地计算排列。方法签名如下所示：

int[][] permute(int n)

例如n = 3，返回值将是：

[[0,1,2],
 [0,2,1],
 [1,0,2],
 [1,2,0],
 [2,0,1],
 [2,1,0]]

您将如何以最有效的方式迭代地执行此操作？我可以递归地执行此操作，但我有兴趣看到许多替代方法来迭代地执行此操作。

4

10 回答 10

23

请参阅 QuickPerm 算法，它是迭代的：http ://www.quickperm.org/

编辑：

为了清楚起见，用 Ruby 重写：

def permute_map(n)
  results = []
  a, p = (0...n).to_a, [0] * n
  i, j = 0, 0
  i = 1
  results << yield(a)
  while i < n
    if p[i] < i
      j = i % 2 * p[i] # If i is odd, then j = p[i], else j = 0
      a[j], a[i] = a[i], a[j] # Swap
      results << yield(a)
      p[i] += 1
      i = 1
    else
      p[i] = 0
      i += 1
    end
  end
  return results
end
于 2010-03-06T01:11:20.093 回答
6

从一个排列到下一个排列的算法与小学加法非常相似 - 当发生溢出时，“carry the one”。

这是我用 C 编写的一个实现：

#include <stdio.h>

//Convenience macro.  Its function should be obvious.
#define swap(a,b) do { \
        typeof(a) __tmp = (a); \
        (a) = (b); \
        (b) = __tmp; \
    } while(0)

void perm_start(unsigned int n[], unsigned int count) {
    unsigned int i;
    for (i=0; i<count; i++)
        n[i] = i;
}

//Returns 0 on wraparound
int perm_next(unsigned int n[], unsigned int count) {
    unsigned int tail, i, j;

    if (count <= 1)
        return 0;

    /* Find all terms at the end that are in reverse order.
       Example: 0 3 (5 4 2 1) (i becomes 2) */
    for (i=count-1; i>0 && n[i-1] >= n[i]; i--);
    tail = i;

    if (tail > 0) {
        /* Find the last item from the tail set greater than
            the last item from the head set, and swap them.
            Example: 0 3* (5 4* 2 1)
            Becomes: 0 4* (5 3* 2 1) */
        for (j=count-1; j>tail && n[j] <= n[tail-1]; j--);

        swap(n[tail-1], n[j]);
    }

    /* Reverse the tail set's order */
    for (i=tail, j=count-1; i<j; i++, j--)
        swap(n[i], n[j]);

    /* If the entire list was in reverse order, tail will be zero. */
    return (tail != 0);
}

int main(void)
{
    #define N 3
    unsigned int perm[N];
    
    perm_start(perm, N);
    do {
        int i;
        for (i = 0; i < N; i++)
            printf("%d ", perm[i]);
        printf("\n");
    } while (perm_next(perm, N));
    
    return 0;
}
于 2010-03-06T01:07:19.883 回答
5

使用 1.9 的 Array#permutation 是一种选择吗？

>> a = [0,1,2].permutation(3).to_a
=> [[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]]
于 2010-03-06T16:50:00.347 回答
4

下面是我在 C# 中的下一个排列算法的泛型版本，与 STL 的函数非常相似next_permutation（但如果它已经是最大可能排列，它不会反转集合，就像 C++ 版本一样）

从理论上讲，它应该适用于任何IList<>IComparables。

static bool NextPermutation<T>(IList<T> a) where T: IComparable
{
    if (a.Count < 2) return false;
    var k = a.Count-2;

    while (k >= 0 && a[k].CompareTo( a[k+1]) >=0) k--;
    if(k<0)return false;

    var l = a.Count - 1;
    while (l > k && a[l].CompareTo(a[k]) <= 0) l--;

    var tmp = a[k];
    a[k] = a[l];
    a[l] = tmp;

    var i = k + 1;
    var j = a.Count - 1;
    while(i<j)
    {
        tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
        i++;
        j--;
    }

    return true;
}

以及演示/测试代码：

var src = "1234".ToCharArray();
do
{
    Console.WriteLine(src);
} 
while (NextPermutation(src));
于 2012-10-07T12:13:18.627 回答
2

I found Joey Adams' version to be the most readable, but I couldn't port it directly to C# because of how C# handles the scoping of for-loop variables. Hence, this is a slightly tweaked version of his code:

/// <summary>
/// Performs an in-place permutation of <paramref name="values"/>, and returns if there 
/// are any more permutations remaining.
/// </summary>
private static bool NextPermutation(int[] values)
{
    if (values.Length == 0)
        throw new ArgumentException("Cannot permutate an empty collection.");

    //Find all terms at the end that are in reverse order.
    //  Example: 0 3 (5 4 2 1) (i becomes 2)
    int tail = values.Length - 1;
    while(tail > 0 && values[tail - 1] >= values[tail])
        tail--;

    if (tail > 0)
    {
        //Find the last item from the tail set greater than the last item from the head 
        //set, and swap them.
        //  Example: 0 3* (5 4* 2 1)
        //  Becomes: 0 4* (5 3* 2 1)
        int index = values.Length - 1;
        while (index > tail && values[index] <= values[tail - 1])
            index--;

        Swap(ref values[tail - 1], ref values[index]);
    }

    //Reverse the tail set's order.
    int limit = (values.Length - tail) / 2;
    for (int index = 0; index < limit; index++)
        Swap(ref values[tail + index], ref values[values.Length - 1 - index]);

    //If the entire list was in reverse order, tail will be zero.
    return (tail != 0);
}

private static void Swap<T>(ref T left, ref T right)
{
    T temp = left;
    left = right;
    right = temp;
}
于 2013-05-09T10:39:01.343 回答
2

这是 C# 中的一个实现，作为扩展方法：

public static IEnumerable<List<T>> Permute<T>(this IList<T> items)
{
    var indexes = Enumerable.Range(0, items.Count).ToArray();

    yield return indexes.Select(idx => items[idx]).ToList();

    var weights = new int[items.Count];
    var idxUpper = 1;
    while (idxUpper < items.Count)
    {
        if (weights[idxUpper] < idxUpper)
        {
            var idxLower = idxUpper % 2 * weights[idxUpper];
            var tmp = indexes[idxLower];
            indexes[idxLower] = indexes[idxUpper];
            indexes[idxUpper] = tmp;
            yield return indexes.Select(idx => items[idx]).ToList();
            weights[idxUpper]++;
            idxUpper = 1;
        }
        else
        {
            weights[idxUpper] = 0;
            idxUpper++;
        }
    }
}

还有一个单元测试：

[TestMethod]
public void Permute()
{
    var ints = new[] { 1, 2, 3 };
    var orderings = ints.Permute().ToList();
    Assert.AreEqual(6, orderings.Count);
    AssertUtil.SequencesAreEqual(new[] { 1, 2, 3 }, orderings[0]);
    AssertUtil.SequencesAreEqual(new[] { 2, 1, 3 }, orderings[1]);
    AssertUtil.SequencesAreEqual(new[] { 3, 1, 2 }, orderings[2]);
    AssertUtil.SequencesAreEqual(new[] { 1, 3, 2 }, orderings[3]);
    AssertUtil.SequencesAreEqual(new[] { 2, 3, 1 }, orderings[4]);
    AssertUtil.SequencesAreEqual(new[] { 3, 2, 1 }, orderings[5]);
}

该方法AssertUtil.SequencesAreEqual是一个自定义测试助手，可以很容易地重新创建。

于 2011-12-05T11:02:17.853 回答
2

我还遇到了另一个答案中引用的 QuickPerm 算法。我还想分享这个答案，因为我看到了一些可以立即做出的改变来缩短它。例如，如果索引数组“p”的初始化稍有不同，则不必在循环之前返回第一个排列。此外，所有这些 while 循环和 if 占用了更多空间。

void permute(char* s, size_t l) {
    int* p = new int[l];
    for (int i = 0; i < l; i++) p[i] = i;
    for (size_t i = 0; i < l; printf("%s\n", s)) {
        std::swap(s[i], s[i % 2 * --p[i]]);
        for (i = 1; p[i] == 0; i++) p[i] = i;
    }
}
于 2014-10-28T01:02:41.837 回答
1

可以迭代调用的递归算法怎么样？如果你真的需要那些东西作为这样的列表（你应该清楚地内联而不是分配一堆毫无意义的内存）。您可以简单地通过其索引即时计算排列。

就像排列是进位加法重新反转尾部（而不是恢复为 0），索引特定排列值是在基数 n 然后 n-1 然后 n-2 中查找数字的数字......通过每次迭代。

public static <T> boolean permutation(List<T> values, int index) {
    return permutation(values, values.size() - 1, index);
}
private static <T> boolean permutation(List<T> values, int n, int index) {
    if ((index == 0) || (n == 0))  return (index == 0);
    Collections.swap(values, n, n-(index % n));
    return permutation(values,n-1,index/n);
}

布尔值返回您的索引值是否超出范围。也就是说，它用完了 n 个值，但仍有剩余的索引。

而且它无法获得超过 12 个对象的所有排列。12！<整数.MAX_VALUE < 13！

——但是，它是如此非常非常漂亮。如果你做错了很多事情可能会有用。

于 2013-05-25T00:18:28.763 回答
1

我已经在 J​​avascript 中实现了该算法。

var all = ["a", "b", "c"];
console.log(permute(all));

function permute(a){
  var i=1,j, temp = "";
  var p = [];
  var n = a.length;
  var output = [];

  output.push(a.slice());
  for(var b=0; b <= n; b++){
    p[b] = b;
  }

  while (i < n){
    p[i]--;
    if(i%2 == 1){
      j = p[i];
    }
    else{
      j = 0;
    }
    temp = a[j];
    a[j] = a[i];
    a[i] = temp;

    i=1;
    while (p[i] === 0){
      p[i] = i;
      i++;
    }
    output.push(a.slice());
  }
  return output;
}
于 2015-05-19T12:03:03.913 回答
0

我已经使用了这里的算法。该页面包含许多有用的信息。

编辑：对不起，那些是递归的。uray 在他的回答中发布了迭代算法的链接。

我创建了一个 PHP 示例。除非您真的需要返回所有结果，否则我只会创建一个迭代类，如下所示：

<?php
class Permutator implements Iterator
{
  private $a, $n, $p, $i, $j, $k;
  private $stop;
    
  public function __construct(array $a)
  {
    $this->a = array_values($a);
    $this->n = count($this->a);
  }
    
  public function current()
  {
    return $this->a;
  }
    
  public function next()
  {
    ++$this->k;
    while ($this->i < $this->n)
    {
      if ($this->p[$this->i] < $this->i)
      {
        $this->j = ($this->i % 2) * $this->p[$this->i];
    
        $tmp = $this->a[$this->j];
        $this->a[$this->j] = $this->a[$this->i];
        $this->a[$this->i] = $tmp;
    
        $this->p[$this->i]++;
        $this->i = 1;
        return;
      }
    
      $this->p[$this->i++] = 0;
    }
    
    $this->stop = true;
  }
    
  public function key()
  {
    return $this->k;
  }
    
  public function valid()
  {
    return !$this->stop;
  }
    
  public function rewind()
  {
    if ($this->n) $this->p = array_fill(0, $this->n, 0);
    $this->stop = $this->n == 0;
    $this->i = 1;
    $this->j = 0;
    $this->k = 0;
  }
    
}
    
foreach (new Permutator(array(1,2,3,4,5)) as $permutation)
{
  var_dump($permutation);
}
?>

请注意，它将每个 PHP 数组视为索引数组。

于 2010-03-06T01:06:15.313 回答