如何找到给定长度的k的排列? [英] How to find permutation of k in a given length?

问题描述

如何找到给定长度的k的排列?

How can I find the permutations of k in a given length?

例如:

单词cat有3个字母:如何找到单词cat中2的所有排列. 结果应为:ac，at，ca，ac等...

The word cat has 3 letters: How can I find all the permutations of 2 in the word cat. Result should be: ac, at, ca, ac, etc...

这不是家庭作业问题. 可以使用任何语言，但更可取的是:C/C ++或C#. 我知道如何为大小LENGTH创建递归，但对于自定义大小却无法创建.

This is not a homework problem. Any language could be used but more preferable: C/C++ or C#. I know how to create the recursion for size LENGTH but not for a custom size.

推荐答案

这是C#中的一个，即使重复字符也可以使用.例如，对于长度为2的排列的香蕉"，它给出:

Here is one in C#, which should work even with repeated characters. For example on "banana" for permutations of length 2 it gives:

ba bn ab aa a nb na nn

ba bn ab aa an nb na nn

基本思想是固定第一个字符，然后形成长度为k-1的所有排列，然后在字符前添加k-1个长度的排列.为了处理重复的字符，我们会跟踪剩余的计数(即可以用于子排列的计数).

The basic idea is to fix the first character, then form all permutations of length k-1, then prepend the character to those k-1 length permutations. To deal with duplicate characters, we keep track of the count left (i.e the ones which can be used for sub-permutations).

不是示例代码，但是应该可以给您带来启发. (如果发现错误，请告诉我，我可以编辑).

Not exemplary code, but should give you the idea. (If you find bugs, let me know and I can edit).

static List<string> Permutations(Dictionary<char, int> input, int length) {
    List<string> permutations = new List<string>();

    List<char> chars = new List<char>(input.Keys);

    // Base case.
    if (length == 0) {
        permutations.Add(string.Empty);
        return permutations;
    }

    foreach (char c in chars) {

        // There are instances of this character left to use.
        if (input[c] > 0) {

            // Use one instance up.
            input[c]--;

            // Find sub-permutations of length length -1.
            List<string> subpermutations = Permutations(input, length - 1);

            // Give back the instance.
            input[c]++;

            foreach (string s in subpermutations) {

                // Prepend the character to be the first character.
                permutations.Add(s.Insert(0,new string(c,1)));

            }
        }
    }

    return permutations;
}

这是我拥有的完整程序，可以使用它:

And here is the full program I have, to use it:

using System;
using System.Collections.Generic;

namespace StackOverflow {

    class Program {
        static void Main(string[] args) {
            List<string> p = Permutations("abracadabra", 3);
            foreach (string s in p) {
                Console.WriteLine(s);
            }
        }

        static List<string> Permutations(string s, int length) {
            Dictionary<char, int> input = new Dictionary<char, int>();
            foreach (char c in s) {
                if (input.ContainsKey(c)) {
                    input[c]++;
                } else {
                    input[c] = 1;
                }
            }
            return Permutations(input, length);
        }

        static List<string> Permutations(Dictionary<char, int> input, 
                                                          int length) {
            List<string> permutations = new List<string>();

            List<char> chars = new List<char>(input.Keys);
            if (length == 0) {
                permutations.Add(string.Empty);
                return permutations;
            }

            foreach (char c in chars) {
                if (input[c] > 0) {
                    input[c]--;
                    List<string> subpermutations = Permutations(input, 
                                                                length - 1);
                    input[c]++;

                    foreach (string s in subpermutations) {
                        permutations.Add(s.Insert(0,new string(c,1)));
                    }
                }
            }

            return permutations;
        }
    }
}

这篇关于如何找到给定长度的k的排列?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！