编程面试中的一项常见任务（尽管不是根据我的面试经验）是获取一个字符串或整数并列出所有可能的排列。
有没有一个例子说明这是如何完成的以及解决这个问题背后的逻辑？
我看过一些代码片段，但它们没有得到很好的评论/解释，因此很难理解。
问问题
199143 次
27 回答
171
首先：它当然闻起来像递归！
既然你也想知道原理，那我尽量用人的语言来解释。我认为递归在大多数时候都很容易。你只需要掌握两个步骤：
- 第一步
- 所有其他步骤（都具有相同的逻辑）
用人类语言：
简而言之：
- 1个元素的排列是一个元素。
- 一组元素的排列是每个元素的列表，与其他元素的每个排列连接。
例子：
如果集合只有一个元素 ->
将其返回。
烫发（一）->一如果集合有两个字符：对于其中的每个元素：返回元素，并添加其余元素的排列，如下所示：
烫发（ab）->
a + 烫发(b) -> ab
b +烫发（a）-> ba
进一步：对于集合中的每个字符：返回一个字符，与 > 集合的其余部分的排列连接
烫发（abc）->
a + perm(bc) --> abc , acb
b + perm(ac) --> bac , bca
c + perm(ab) --> 出租车，cba
烫发（abc...z）-->
a + 烫发(...), b + 烫发(....)
....
我在http://www.programmersheaven.com/mb/Algorithms/369713/369713/permutation-algorithm-help/上找到了伪代码：
makePermutations(permutation) {
if (length permutation < required length) {
for (i = min digit to max digit) {
if (i not in permutation) {
makePermutations(permutation+i)
}
}
}
else {
add permutation to list
}
}
C＃
好的，还有更详细的内容（因为它被标记为 c#），来自http://radio.weblogs.com/0111551/stories/2002/10/14/permutations.html：相当冗长，但我决定复制它无论如何，所以帖子不依赖于原件。
该函数接受一个字符串，并写下该确切字符串的所有可能排列，例如，如果提供了“ABC”，则应该溢出：
ABC、ACB、BAC、BCA、CAB、CBA。
代码：
class Program
{
private static void Swap(ref char a, ref char b)
{
if (a == b) return;
var temp = a;
a = b;
b = temp;
}
public static void GetPer(char[] list)
{
int x = list.Length - 1;
GetPer(list, 0, x);
}
private static void GetPer(char[] list, int k, int m)
{
if (k == m)
{
Console.Write(list);
}
else
for (int i = k; i <= m; i++)
{
Swap(ref list[k], ref list[i]);
GetPer(list, k + 1, m);
Swap(ref list[k], ref list[i]);
}
}
static void Main()
{
string str = "sagiv";
char[] arr = str.ToCharArray();
GetPer(arr);
}
}
于 2009-04-16T13:22:28.210 回答
95
如果允许使用 LINQ，则只需两行代码。请在此处查看我的答案。
编辑
这是我的通用函数，它可以从 T 列表中返回所有排列（不是组合）：
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 }));
}
例子：
IEnumerable<IEnumerable<int>> result =
GetPermutations(Enumerable.Range(1, 3), 3);
输出 - 整数列表列表：
{1,2,3} {1,3,2} {2,1,3} {2,3,1} {3,1,2} {3,2,1}
由于此函数使用 LINQ，因此它需要 .net 3.5 或更高版本。
于 2012-05-17T04:54:32.867 回答
41
在这里，我找到了解决方案。它是用 Java 编写的，但我已将其转换为 C#。我希望它会帮助你。
这是 C# 中的代码：
static void Main(string[] args)
{
string str = "ABC";
char[] charArry = str.ToCharArray();
Permute(charArry, 0, 2);
Console.ReadKey();
}
static void Permute(char[] arry, int i, int n)
{
int j;
if (i==n)
Console.WriteLine(arry);
else
{
for(j = i; j <=n; j++)
{
Swap(ref arry[i],ref arry[j]);
Permute(arry,i+1,n);
Swap(ref arry[i], ref arry[j]); //backtrack
}
}
}
static void Swap(ref char a, ref char b)
{
char tmp;
tmp = a;
a=b;
b = tmp;
}
于 2014-02-18T03:11:27.690 回答
22
递归不是必需的，这里有关于这个解决方案的好信息。
var values1 = new[] { 1, 2, 3, 4, 5 };
foreach (var permutation in values1.GetPermutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
var values2 = new[] { 'a', 'b', 'c', 'd', 'e' };
foreach (var permutation in values2.GetPermutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
Console.ReadLine();
我多年来一直使用这种算法，它具有O(N) 时间和空间复杂度来计算每个排列。
public static class SomeExtensions
{
public static IEnumerable<IEnumerable<T>> GetPermutations<T>(this IEnumerable<T> enumerable)
{
var array = enumerable as T[] ?? enumerable.ToArray();
var factorials = Enumerable.Range(0, array.Length + 1)
.Select(Factorial)
.ToArray();
for (var i = 0L; i < factorials[array.Length]; i++)
{
var sequence = GenerateSequence(i, array.Length - 1, factorials);
yield return GeneratePermutation(array, sequence);
}
}
private static IEnumerable<T> GeneratePermutation<T>(T[] array, IReadOnlyList<int> sequence)
{
var clone = (T[]) array.Clone();
for (int i = 0; i < clone.Length - 1; i++)
{
Swap(ref clone[i], ref clone[i + sequence[i]]);
}
return clone;
}
private static int[] GenerateSequence(long number, int size, IReadOnlyList<long> factorials)
{
var sequence = new int[size];
for (var j = 0; j < sequence.Length; j++)
{
var facto = factorials[sequence.Length - j];
sequence[j] = (int)(number / facto);
number = (int)(number % facto);
}
return sequence;
}
static void Swap<T>(ref T a, ref T b)
{
T temp = a;
a = b;
b = temp;
}
private static long Factorial(int n)
{
long result = n;
for (int i = 1; i < n; i++)
{
result = result * i;
}
return result;
}
}
于 2015-09-12T23:47:19.577 回答
13
class Program
{
public static void Main(string[] args)
{
Permutation("abc");
}
static void Permutation(string rest, string prefix = "")
{
if (string.IsNullOrEmpty(rest)) Console.WriteLine(prefix);
// Each letter has a chance to be permutated
for (int i = 0; i < rest.Length; i++)
{
Permutation(rest.Remove(i, 1), prefix + rest[i]);
}
}
}
于 2018-05-07T17:23:21.683 回答
11
C# 中稍作修改的版本，可在 ANY 类型的数组中产生所需的排列。
// USAGE: create an array of any type, and call Permutations()
var vals = new[] {"a", "bb", "ccc"};
foreach (var v in Permutations(vals))
Console.WriteLine(string.Join(",", v)); // Print values separated by comma
public static IEnumerable<T[]> Permutations<T>(T[] values, int fromInd = 0)
{
if (fromInd + 1 == values.Length)
yield return values;
else
{
foreach (var v in Permutations(values, fromInd + 1))
yield return v;
for (var i = fromInd + 1; i < values.Length; i++)
{
SwapValues(values, fromInd, i);
foreach (var v in Permutations(values, fromInd + 1))
yield return v;
SwapValues(values, fromInd, i);
}
}
}
private static void SwapValues<T>(T[] values, int pos1, int pos2)
{
if (pos1 != pos2)
{
T tmp = values[pos1];
values[pos1] = values[pos2];
values[pos2] = tmp;
}
}
于 2012-10-23T00:44:50.447 回答
9
首先，集合有排列，而不是字符串或整数，所以我假设你的意思是“字符串中的字符集”。
请注意，一组大小为 n 的集合有 n! n-排列。
以下伪代码（来自 Wikipedia），以 k = 1...n 调用！将给出所有排列：
function permutation(k, s) {
for j = 2 to length(s) {
swap s[(k mod j) + 1] with s[j]; // note that our array is indexed starting at 1
k := k / j; // integer division cuts off the remainder
}
return s;
}
这是等效的 Python 代码（用于基于 0 的数组索引）：
def permutation(k, s):
r = s[:]
for j in range(2, len(s)+1):
r[j-1], r[k%j] = r[k%j], r[j-1]
k = k/j+1
return r
于 2009-04-16T13:25:08.173 回答
7
我喜欢FBryant87方法，因为它很简单。不幸的是，它确实像许多其他“解决方案”一样不提供所有排列或例如整数，如果它包含多次相同的数字。以656123为例。该行：
var tail = chars.Except(new List<char>(){c});
使用 except 将导致所有出现的事件都被删除，即当 c = 6 时，两个数字被删除，我们只剩下例如 5123。由于我尝试的解决方案都没有解决这个问题，我决定尝试通过 FBryant87 自己解决它代码作为基础。这就是我想出的：
private static List<string> FindPermutations(string set)
{
var output = new List<string>();
if (set.Length == 1)
{
output.Add(set);
}
else
{
foreach (var c in set)
{
// Remove one occurrence of the char (not all)
var tail = set.Remove(set.IndexOf(c), 1);
foreach (var tailPerms in FindPermutations(tail))
{
output.Add(c + tailPerms);
}
}
}
return output;
}
我只是使用 .Remove 和 .IndexOf 删除第一个发现的事件。似乎至少可以按我的预期工作。我相信它可以变得更聪明。
不过需要注意的一点是：结果列表可能包含重复项，因此请确保您要么让方法返回例如 HashSet，要么在返回后使用您喜欢的任何方法删除重复项。
于 2015-05-18T20:23:29.230 回答
5
这是一个纯函数式 F# 实现：
let factorial i =
let rec fact n x =
match n with
| 0 -> 1
| 1 -> x
| _ -> fact (n-1) (x*n)
fact i 1
let swap (arr:'a array) i j = [| for k in 0..(arr.Length-1) -> if k = i then arr.[j] elif k = j then arr.[i] else arr.[k] |]
let rec permutation (k:int,j:int) (r:'a array) =
if j = (r.Length + 1) then r
else permutation (k/j+1, j+1) (swap r (j-1) (k%j))
let permutations (source:'a array) = seq { for k = 0 to (source |> Array.length |> factorial) - 1 do yield permutation (k,2) source }
通过更改交换以利用 CLR 数组的可变特性可以大大提高性能，但是这种实现对于源数组是线程安全的，并且在某些情况下可能是可取的。此外，对于超过 16 个元素的数组，必须将 int 替换为具有更高/任意精度的类型，因为阶乘 17 会导致 int32 溢出。
于 2009-04-16T14:46:52.533 回答
5
这是一个易于理解的置换函数，可将字符串和整数作为输入。有了这个你甚至可以设置你的输出长度（在正常情况下它等于输入长度）
细绳
static ICollection<string> result;
public static ICollection<string> GetAllPermutations(string str, int outputLength)
{
result = new List<string>();
MakePermutations(str.ToCharArray(), string.Empty, outputLength);
return result;
}
private static void MakePermutations(
char[] possibleArray,//all chars extracted from input
string permutation,
int outputLength//the length of output)
{
if (permutation.Length < outputLength)
{
for (int i = 0; i < possibleArray.Length; i++)
{
var tempList = possibleArray.ToList<char>();
tempList.RemoveAt(i);
MakePermutations(tempList.ToArray(),
string.Concat(permutation, possibleArray[i]), outputLength);
}
}
else if (!result.Contains(permutation))
result.Add(permutation);
}
对于Integer，只需更改调用者方法，MakePermutations()保持不变：
public static ICollection<int> GetAllPermutations(int input, int outputLength)
{
result = new List<string>();
MakePermutations(input.ToString().ToCharArray(), string.Empty, outputLength);
return result.Select(m => int.Parse(m)).ToList<int>();
}
示例 1：GetAllPermutations("abc",3); “abc” “acb” “bac” “bca” “cab” “cba”
示例 2：GetAllPermutations("abcd",2); “ab” “ac” “ad” “ba” “bc” “bd” “ca” “cb” “cd” “da” “db” “dc”
示例 3：GetAllPermutations(486,2)；48 46 84 86 64 68
于 2016-07-04T00:50:33.863 回答
5
这是使用递归的 c# 中的简单解决方案，
void Main()
{
string word = "abc";
WordPermuatation("",word);
}
void WordPermuatation(string prefix, string word)
{
int n = word.Length;
if (n == 0) { Console.WriteLine(prefix); }
else
{
for (int i = 0; i < n; i++)
{
WordPermuatation(prefix + word[i],word.Substring(0, i) + word.Substring(i + 1, n - (i+1)));
}
}
}
于 2016-10-20T15:16:32.633 回答
3
基于@Peter 的解决方案，这里有一个版本，它声明了一个简单的 LINQ 风格的Permutations()扩展方法，适用于任何IEnumerable<T>.
用法（以字符串字符为例）：
foreach (var permutation in "abc".Permutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
输出：
a, b, c
a, c, b
b, a, c
b, c, a
c, b, a
c, a, b
或在任何其他集合类型上：
foreach (var permutation in (new[] { "Apples", "Oranges", "Pears"}).Permutations())
{
Console.WriteLine(string.Join(", ", permutation));
}
输出：
Apples, Oranges, Pears
Apples, Pears, Oranges
Oranges, Apples, Pears
Oranges, Pears, Apples
Pears, Oranges, Apples
Pears, Apples, Oranges
using System;
using System.Collections.Generic;
using System.Linq;
public static class PermutationExtension
{
public static IEnumerable<T[]> Permutations<T>(this IEnumerable<T> source)
{
var sourceArray = source.ToArray();
var results = new List<T[]>();
Permute(sourceArray, 0, sourceArray.Length - 1, results);
return results;
}
private static void Swap<T>(ref T a, ref T b)
{
T tmp = a;
a = b;
b = tmp;
}
private static void Permute<T>(T[] elements, int recursionDepth, int maxDepth, ICollection<T[]> results)
{
if (recursionDepth == maxDepth)
{
results.Add(elements.ToArray());
return;
}
for (var i = recursionDepth; i <= maxDepth; i++)
{
Swap(ref elements[recursionDepth], ref elements[i]);
Permute(elements, recursionDepth + 1, maxDepth, results);
Swap(ref elements[recursionDepth], ref elements[i]);
}
}
}
于 2019-11-12T21:15:56.540 回答
2
这是将打印所有排列的函数。这个函数实现了peter解释的逻辑。
public class Permutation
{
//http://www.java2s.com/Tutorial/Java/0100__Class-Definition/RecursivemethodtofindallpermutationsofaString.htm
public static void permuteString(String beginningString, String endingString)
{
if (endingString.Length <= 1)
Console.WriteLine(beginningString + endingString);
else
for (int i = 0; i < endingString.Length; i++)
{
String newString = endingString.Substring(0, i) + endingString.Substring(i + 1);
permuteString(beginningString + endingString.ElementAt(i), newString);
}
}
}
static void Main(string[] args)
{
Permutation.permuteString(String.Empty, "abc");
Console.ReadLine();
}
于 2011-12-19T20:09:19.527 回答
2
下面是我对 permutation 的实现。不要介意变量名，因为我这样做是为了好玩:)
class combinations
{
static void Main()
{
string choice = "y";
do
{
try
{
Console.WriteLine("Enter word :");
string abc = Console.ReadLine().ToString();
Console.WriteLine("Combinatins for word :");
List<string> final = comb(abc);
int count = 1;
foreach (string s in final)
{
Console.WriteLine("{0} --> {1}", count++, s);
}
Console.WriteLine("Do you wish to continue(y/n)?");
choice = Console.ReadLine().ToString();
}
catch (Exception exc)
{
Console.WriteLine(exc);
}
} while (choice == "y" || choice == "Y");
}
static string swap(string test)
{
return swap(0, 1, test);
}
static List<string> comb(string test)
{
List<string> sec = new List<string>();
List<string> first = new List<string>();
if (test.Length == 1) first.Add(test);
else if (test.Length == 2) { first.Add(test); first.Add(swap(test)); }
else if (test.Length > 2)
{
sec = generateWords(test);
foreach (string s in sec)
{
string init = s.Substring(0, 1);
string restOfbody = s.Substring(1, s.Length - 1);
List<string> third = comb(restOfbody);
foreach (string s1 in third)
{
if (!first.Contains(init + s1)) first.Add(init + s1);
}
}
}
return first;
}
static string ShiftBack(string abc)
{
char[] arr = abc.ToCharArray();
char temp = arr[0];
string wrd = string.Empty;
for (int i = 1; i < arr.Length; i++)
{
wrd += arr[i];
}
wrd += temp;
return wrd;
}
static List<string> generateWords(string test)
{
List<string> final = new List<string>();
if (test.Length == 1)
final.Add(test);
else
{
final.Add(test);
string holdString = test;
while (final.Count < test.Length)
{
holdString = ShiftBack(holdString);
final.Add(holdString);
}
}
return final;
}
static string swap(int currentPosition, int targetPosition, string temp)
{
char[] arr = temp.ToCharArray();
char t = arr[currentPosition];
arr[currentPosition] = arr[targetPosition];
arr[targetPosition] = t;
string word = string.Empty;
for (int i = 0; i < arr.Length; i++)
{
word += arr[i];
}
return word;
}
}
于 2012-09-22T08:29:58.610 回答
2
这是我写的一个高级示例，它说明了彼得给出的人类语言解释：
public List<string> FindPermutations(string input)
{
if (input.Length == 1)
return new List<string> { input };
var perms = new List<string>();
foreach (var c in input)
{
var others = input.Remove(input.IndexOf(c), 1);
perms.AddRange(FindPermutations(others).Select(perm => c + perm));
}
return perms;
}
于 2014-08-19T15:08:37.000 回答
1
这是我的解决方案，我很容易理解
class ClassicPermutationProblem
{
ClassicPermutationProblem() { }
private static void PopulatePosition<T>(List<List<T>> finalList, List<T> list, List<T> temp, int position)
{
foreach (T element in list)
{
List<T> currentTemp = temp.ToList();
if (!currentTemp.Contains(element))
currentTemp.Add(element);
else
continue;
if (position == list.Count)
finalList.Add(currentTemp);
else
PopulatePosition(finalList, list, currentTemp, position + 1);
}
}
public static List<List<int>> GetPermutations(List<int> list)
{
List<List<int>> results = new List<List<int>>();
PopulatePosition(results, list, new List<int>(), 1);
return results;
}
}
static void Main(string[] args)
{
List<List<int>> results = ClassicPermutationProblem.GetPermutations(new List<int>() { 1, 2, 3 });
}
于 2015-08-16T17:28:09.103 回答
1
如果性能和内存是一个问题，我建议这种非常有效的实现。根据Wikipedia 中的 Heap 算法，应该是最快的。希望它能满足您的需要:-)！
就像将此与 10 的 Linq 实现进行比较！（包括代码）：
- 这：36288000 个项目在 235 毫秒内
Linq：50051 毫秒内 36288000 个项目
using System; using System.Collections.Generic; using System.Diagnostics; using System.Linq; using System.Runtime.CompilerServices; using System.Text; 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 /// </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) => { Debug.Print(String.Join("", vals)); return false; }); int[] values = new int[] { 0, 1, 2, 4 }; Debug.Print("Non Linq"); ForAllPermutation(values, (vals) => { Debug.Print(String.Join("", vals)); return false; }); Debug.Print("Linq"); foreach(var v in GetPermutations(values, values.Length)) { Debug.Print(String.Join("", v)); } // Performance int count = 0; values = new int[10]; for(int n = 0; n < values.Length; n++) { values[n] = n; } Stopwatch stopWatch = new Stopwatch(); stopWatch.Reset(); stopWatch.Start(); ForAllPermutation(values, (vals) => { foreach(var v in vals) { count++; } return false; }); stopWatch.Stop(); Debug.Print($"Non Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs"); count = 0; stopWatch.Reset(); stopWatch.Start(); foreach (var vals in GetPermutations(values, values.Length)) { foreach (var v in vals) { count++; } } stopWatch.Stop(); Debug.Print($"Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs"); } } }
于 2016-04-14T21:20:51.897 回答
1
这是我在 JavaScript (NodeJS) 中的解决方案。主要思想是我们一次取一个元素，从字符串中“删除”它，改变其余字符，然后在前面插入元素。
function perms (string) {
if (string.length == 0) {
return [];
}
if (string.length == 1) {
return [string];
}
var list = [];
for(var i = 0; i < string.length; i++) {
var invariant = string[i];
var rest = string.substr(0, i) + string.substr(i + 1);
var newPerms = perms(rest);
for (var j = 0; j < newPerms.length; j++) {
list.push(invariant + newPerms[j]);
}
}
return list;
}
module.exports = perms;
以下是测试：
require('should');
var permutations = require('../src/perms');
describe('permutations', function () {
it('should permute ""', function () {
permutations('').should.eql([]);
})
it('should permute "1"', function () {
permutations('1').should.eql(['1']);
})
it('should permute "12"', function () {
permutations('12').should.eql(['12', '21']);
})
it('should permute "123"', function () {
var expected = ['123', '132', '321', '213', '231', '312'];
var actual = permutations('123');
expected.forEach(function (e) {
actual.should.containEql(e);
})
})
it('should permute "1234"', function () {
// Wolfram Alpha FTW!
var expected = ['1234', '1243', '1324', '1342', '1423', '1432', '2134', '2143', '2314', '2341', '2413', '2431', '3124', '3142', '3214', '3241', '3412', '3421', '4123', '4132'];
var actual = permutations('1234');
expected.forEach(function (e) {
actual.should.containEql(e);
})
})
})
于 2017-01-13T02:55:48.740 回答
1
这是我能想到的最简单的解决方案：
let rec distribute e = function
| [] -> [[e]]
| x::xs' as xs -> (e::xs)::[for xs in distribute e xs' -> x::xs]
let permute xs = Seq.fold (fun ps x -> List.collect (distribute x) ps) [[]] xs
该distribute函数接受一个新元素e和一个n-element 列表，并返回一个列表列表，n+1每个列表都e插入到不同的位置。例如，10在列表的四个可能位置中的每一个插入[1;2;3]：
> distribute 10 [1..3];;
val it : int list list =
[[10; 1; 2; 3]; [1; 10; 2; 3]; [1; 2; 10; 3]; [1; 2; 3; 10]]
该permute函数依次折叠每个元素，分布在迄今为止累积的排列上，最终形成所有排列。例如，列表的 6 个排列[1;2;3]：
> permute [1;2;3];;
val it : int list list =
[[3; 2; 1]; [2; 3; 1]; [2; 1; 3]; [3; 1; 2]; [1; 3; 2]; [1; 2; 3]]
将 更改fold为 ascan以保留中间累加器可以阐明排列是如何一次生成一个元素的：
> Seq.scan (fun ps x -> List.collect (distribute x) ps) [[]] [1..3];;
val it : seq<int list list> =
seq
[[[]]; [[1]]; [[2; 1]; [1; 2]];
[[3; 2; 1]; [2; 3; 1]; [2; 1; 3]; [3; 1; 2]; [1; 3; 2]; [1; 2; 3]]]
于 2017-02-22T19:00:50.487 回答
1
列出字符串的排列。当字符重复时避免重复：
using System;
using System.Collections;
class Permutation{
static IEnumerable Permutations(string word){
if (word == null || word.Length <= 1) {
yield return word;
yield break;
}
char firstChar = word[0];
foreach( string subPermute in Permutations (word.Substring (1)) ) {
int indexOfFirstChar = subPermute.IndexOf (firstChar);
if (indexOfFirstChar == -1) indexOfFirstChar = subPermute.Length;
for( int index = 0; index <= indexOfFirstChar; index++ )
yield return subPermute.Insert (index, new string (firstChar, 1));
}
}
static void Main(){
foreach( var permutation in Permutations ("aab") )
Console.WriteLine (permutation);
}
}
于 2017-04-24T07:39:36.833 回答
1
//Generic C# Method
private static List<T[]> GetPerms<T>(T[] input, int startIndex = 0)
{
var perms = new List<T[]>();
var l = input.Length - 1;
if (l == startIndex)
perms.Add(input);
else
{
for (int i = startIndex; i <= l; i++)
{
var copy = input.ToArray(); //make copy
var temp = copy[startIndex];
copy[startIndex] = copy[i];
copy[i] = temp;
perms.AddRange(GetPerms(copy, startIndex + 1));
}
}
return perms;
}
//usages
char[] charArray = new char[] { 'A', 'B', 'C' };
var charPerms = GetPerms(charArray);
string[] stringArray = new string[] { "Orange", "Mango", "Apple" };
var stringPerms = GetPerms(stringArray);
int[] intArray = new int[] { 1, 2, 3 };
var intPerms = GetPerms(intArray);
于 2017-12-31T11:09:36.700 回答
0
这是递归打印所有排列的函数。
public void Permutations(string input, StringBuilder sb)
{
if (sb.Length == input.Length)
{
Console.WriteLine(sb.ToString());
return;
}
char[] inChar = input.ToCharArray();
for (int i = 0; i < input.Length; i++)
{
if (!sb.ToString().Contains(inChar[i]))
{
sb.Append(inChar[i]);
Permutations(input, sb);
RemoveChar(sb, inChar[i]);
}
}
}
private bool RemoveChar(StringBuilder input, char toRemove)
{
int index = input.ToString().IndexOf(toRemove);
if (index >= 0)
{
input.Remove(index, 1);
return true;
}
return false;
}
于 2013-08-12T07:28:53.033 回答
0
class Permutation
{
public static List<string> Permutate(string seed, List<string> lstsList)
{
loopCounter = 0;
// string s="\w{0,2}";
var lstStrs = PermuateRecursive(seed);
Trace.WriteLine("Loop counter :" + loopCounter);
return lstStrs;
}
// Recursive function to find permutation
private static List<string> PermuateRecursive(string seed)
{
List<string> lstStrs = new List<string>();
if (seed.Length > 2)
{
for (int i = 0; i < seed.Length; i++)
{
str = Swap(seed, 0, i);
PermuateRecursive(str.Substring(1, str.Length - 1)).ForEach(
s =>
{
lstStrs.Add(str[0] + s);
loopCounter++;
});
;
}
}
else
{
lstStrs.Add(seed);
lstStrs.Add(Swap(seed, 0, 1));
}
return lstStrs;
}
//Loop counter variable to count total number of loop execution in various functions
private static int loopCounter = 0;
//Non recursive version of permuation function
public static List<string> Permutate(string seed)
{
loopCounter = 0;
List<string> strList = new List<string>();
strList.Add(seed);
for (int i = 0; i < seed.Length; i++)
{
int count = strList.Count;
for (int j = i + 1; j < seed.Length; j++)
{
for (int k = 0; k < count; k++)
{
strList.Add(Swap(strList[k], i, j));
loopCounter++;
}
}
}
Trace.WriteLine("Loop counter :" + loopCounter);
return strList;
}
private static string Swap(string seed, int p, int p2)
{
Char[] chars = seed.ToCharArray();
char temp = chars[p2];
chars[p2] = chars[p];
chars[p] = temp;
return new string(chars);
}
}
于 2014-02-23T03:35:51.603 回答
0
这是一个稍微简化的 C# 答案。
public static void StringPermutationsDemo()
{
strBldr = new StringBuilder();
string result = Permute("ABCD".ToCharArray(), 0);
MessageBox.Show(result);
}
static string Permute(char[] elementsList, int startIndex)
{
if (startIndex == elementsList.Length)
{
foreach (char element in elementsList)
{
strBldr.Append(" " + element);
}
strBldr.AppendLine("");
}
else
{
for (int tempIndex = startIndex; tempIndex <= elementsList.Length - 1; tempIndex++)
{
Swap(ref elementsList[startIndex], ref elementsList[tempIndex]);
Permute(elementsList, (startIndex + 1));
Swap(ref elementsList[startIndex], ref elementsList[tempIndex]);
}
}
return strBldr.ToString();
}
static void Swap(ref char Char1, ref char Char2)
{
char tempElement = Char1;
Char1 = Char2;
Char2 = tempElement;
}
输出：
1 2 3
1 3 2
2 1 3
2 3 1
3 2 1
3 1 2
于 2014-04-18T04:22:42.777 回答
0
这是提到的算法的另一种实现。
public class Program
{
public static void Main(string[] args)
{
string str = "abcefgh";
var astr = new Permutation().GenerateFor(str);
Console.WriteLine(astr.Length);
foreach(var a in astr)
{
Console.WriteLine(a);
}
//a.ForEach(Console.WriteLine);
}
}
class Permutation
{
public string[] GenerateFor(string s)
{
if(s.Length == 1)
{
return new []{s};
}
else if(s.Length == 2)
{
return new []{s[1].ToString()+s[0].ToString(),s[0].ToString()+s[1].ToString()};
}
var comb = new List<string>();
foreach(var c in s)
{
string cStr = c.ToString();
var sToProcess = s.Replace(cStr,"");
if (!string.IsNullOrEmpty(sToProcess) && sToProcess.Length>0)
{
var conCatStr = GenerateFor(sToProcess);
foreach(var a in conCatStr)
{
comb.Add(c.ToString()+a);
}
}
}
return comb.ToArray();
}
}
于 2016-07-05T21:40:47.630 回答
0
基于鹏阳 答案的基础/修订
灵感来自permutations-in-javascript
c#版本
FunctionalPermutations应该是这个
static IEnumerable<IEnumerable<T>> FunctionalPermutations<T>(IEnumerable<T> elements, int length)
{
if (length < 2) return elements.Select(t => new T[] { t });
/* Pengyang answser..
return _recur_(list, length - 1).SelectMany(t => list.Where(e => !t.Contains(e)),(t1, t2) => t1.Concat(new T[] { t2 }));
*/
return elements.SelectMany((element_i, i) =>
FunctionalPermutations(elements.Take(i).Concat(elements.Skip(i + 1)), length - 1)
.Select(sub_ei => new[] { element_i }.Concat(sub_ei)));
}
于 2020-10-10T10:59:39.237 回答
0
我希望这就足够了：
using System;
public class Program
{
public static void Main()
{
//Example using word cat
permute("cat");
}
static void permute(string word){
for(int i=0; i < word.Length; i++){
char start = word[0];
for(int j=1; j < word.Length; j++){
string left = word.Substring(1,j-1);
string right = word.Substring(j);
Console.WriteLine(start+right+left);
}
if(i+1 < word.Length){
word = wordChange(word, i + 1);
}
}
}
static string wordChange(string word, int index){
string newWord = "";
for(int i=0; i<word.Length; i++){
if(i== 0)
newWord += word[index];
else if(i== index)
newWord += word[0];
else
newWord += word[i];
}
return newWord;
}
输出：
cat
cta
act
atc
tca
tac
于 2021-01-22T14:13:13.157 回答