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比你的正常排列还要多

permutation .net c#

2

oshirowanen · 技术社区 · 7 年前

following example 用于生成置换的goes:

namespace ConsoleApp1
{
    class Program
    {
        public static void Main()
        {
            int n, i;
            formPermut test = new formPermut();
            int[] arr1 = new int[5];

            Console.WriteLine("\n\n Recursion : Generate all possible permutations of an array :");
            Console.WriteLine("------------------------------------------------------------------");

            Console.Write(" Input the number of elements to store in the array [maximum 5 digits ] :");
            n = Convert.ToInt32(Console.ReadLine());
            Console.Write(" Input {0} number of elements in the array :\n", n);
            for (i = 0; i < n; i++)
            {
                Console.Write(" element - {0} : ", i);
                arr1[i] = Convert.ToInt32(Console.ReadLine());
            }

            Console.Write("\n The Permutations with a combination of {0} digits are : \n", n);
            test.prnPermut(arr1, 0, n - 1);
            Console.Write("\n\n");
            Console.ReadKey();
        }

        class formPermut
        {
            public void swapTwoNumber(ref int a, ref int b)
            {
                int temp = a;
                a = b;
                b = temp;
            }
            public void prnPermut(int[] list, int k, int m)
            {
                int i;
                if (k == m)
                {
                    for (i = 0; i <= m; i++)
                        Console.Write("{0}", list[i]);
                    Console.Write(" ");
                }
                else
                    for (i = k; i <= m; i++)
                    {
                        swapTwoNumber(ref list[k], ref list[i]);
                        prnPermut(list, k + 1, m);
                        swapTwoNumber(ref list[k], ref list[i]);
                    }
            }
        }
    }
}

1 和 2 ,上述代码将返回以下结果 12 和 21

1. 2. 21

和 2. 和 3 ,在返回时:

123 132 213 231 321 312

有人知道我该怎么做吗?

例如:

1 2 3 12 13 21 23 31 32 123 132 213 321 312

我的最终目标是能够做同样的事情,但是使用字符串,所以如果输入是 one 和 two ,则输出为:

onetwo 二 twoone

和 二 和 three

产出将是:

one two three onetwo onethree twoone twothree threeone threetwo onetwothree onethreetwo twoonethree threetwoone threeonetwo 假设我没有遗漏任何组合。

1 回复 | 直到 7 年前

1

Elliveny 7 年前

这似乎能解决问题?

测试用例:

输入:1,2给出。。。

2 1 21 12

输入1,2,3给出。。。

3 2 32 23 1 31 13 21 12 321 312 231 213 123 132

three two threetwo twothree one threeone onethree twoone onetwo threetwoone threeonetwo twothreeone twoonethree onetwothree onethreetwo

根据该代码:

class Program
{
    public static void Main()
    {
        formPermut test = new formPermut();
        test.prnPermutWithSubsets(new object[] { 1, 2 });
        Console.WriteLine();
        test.prnPermutWithSubsets(new object[] { 1, 2, 3 });
        Console.WriteLine();
        test.prnPermutWithSubsets(new string[] { "one", "two", "three" });
        Console.WriteLine();
        return;
    }

    class formPermut
    {
        private void swapTwoNumber(ref object a, ref object b)
        {
            object temp = a;
            a = b;
            b = temp;
        }
        public void prnPermutWithSubsets(object[] list)
        {
            for (int i = 0; i < Math.Pow(2, list.Length); i++)
            {
                Stack<object> combination = new Stack<object>();
                for (int j = 0; j < list.Length; j++)
                {
                    if ((i & (1 << (list.Length - j - 1))) != 0)
                    {
                        combination.Push(list[j]);
                    }
                }
                this.prnPermut(combination.ToArray(), 0, combination.Count() - 1);
            }
        }

        public void prnPermut(object[] list, int k, int m)
        {
            int i;
            if (k == m)
            {
                for (i = 0; i <= m; i++)
                    Console.Write("{0}", list[i]);
                Console.Write(" ");
            }
            else
                for (i = k; i <= m; i++)
                {
                    swapTwoNumber(ref list[k], ref list[i]);
                    prnPermut(list, k + 1, m);
                    swapTwoNumber(ref list[k], ref list[i]);
                }
        }
    }
}