如何找到二进制数的所有选项[重复]答案

【问题标题】：How to find all of the options for binary number [duplicate]如何找到二进制数的所有选项[重复]
【发布时间】：2015-02-06 05:10:34
【问题描述】：

我有一个包含二进制数的字符串数组： {“xx0x”、“110x”、“100x”、“010x”、“x01x”、“1010”、“0011”、“0111”、“1111”、“xxxx”、“0001”、“1010”、“ 0110" };

“x”可以是 0 或 1。我需要创建一个包含所有存在可能性的新数组。

我想过创建一个包含所有选项的数组然后进行比较，但我不确定如何进行比较。

【问题讨论】：

我已经编辑了你的标题。请参阅“Should questions include “tags” in their titles?”，其中的共识是“不，他们不应该”。
对我来说听起来像是一个计划。只是不要在掩码中有x 时比较位置。但请注意，xxxx 将涵盖所有可能的结果，那么这是一个错字吗？
我仍然不确定如何进行比较，如何忽略部分字符？是的，xxxx 给了我所有的选项，但是我在数字的开头有额外的 4 个常量数字，我不需要检查，这会有所不同:)
发布您的代码，不要忽略您无法弄清楚的部分，我们可能会从那里帮助您。
这是完整的数组：string[] InputsToSOD = new string[] { "0000xx0x", "0000110x", "0000100x", "0000010x", "0000x01x", "00001010", "00000011 ", "00000111", "00001111", "1000xxxx", "10000001", "10001010", "10000110" };

标签： c# string binary

【解决方案1】：

这是一个枚举所有可能值的代码。

           string[] patterns = { "xx0x", "110x", "100x", "010x", "x01x", "1010", "0011", "0111", "1111", "xxxx", "0001", "1010", "0110" };

        // - Loop for each input pattern
        foreach (string pattern in patterns)
        {
            // - Count the varying characters and memorize their position
            int wildcardsCount = pattern.Count(x => x == 'x');
            int[] positions = new int[wildcardsCount];
            int index = 0;
            int position = 0;
            foreach (char c in pattern)
            {
                if (c == 'x')
                    positions[index++] = position;
                position++;
            }

            // - Loop in all possible values taken by missing bits
            int count = (int)Math.Pow(2, wildcardsCount);
            for (int currentValue = 0; currentValue < count; currentValue++)
            {
                // - Prepare the binary string
                string currentValueStr = Convert.ToString(currentValue, 2).PadLeft(wildcardsCount, '0');

                // - Replace each character with known wildcard positions
                char[] tmpOutput = pattern.ToCharArray();
                int charIndex = 0;
                for (charIndex = 0; charIndex < wildcardsCount; charIndex++)
                    tmpOutput[positions[charIndex]] = currentValueStr[charIndex];
                string output = new string(tmpOutput);

                // - Here we are
                Debug.WriteLine(pattern + " (" + (currentValue + 1).ToString() + "/" + count.ToString() + ") => " + output);
            }
        }

输出：

xx0x (1/8) => 0000
xx0x (2/8) => 0001
xx0x (3/8) => 0100
xx0x (4/8) => 0101
xx0x (5/8) => 1000
xx0x (6/8) => 1001
xx0x (7/8) => 1100
xx0x (8/8) => 1101
110x (1/2) => 1100
110x (2/2) => 1101
100x (1/2) => 1000
100x (2/2) => 1001
010x (1/2) => 0100
010x (2/2) => 0101
x01x (1/4) => 0010
x01x (2/4) => 0011
x01x (3/4) => 1010
x01x (4/4) => 1011
1010 (1/1) => 1010
0011 (1/1) => 0011
0111 (1/1) => 0111
1111 (1/1) => 1111
xxxx (1/16) => 0000
xxxx (2/16) => 0001
xxxx (3/16) => 0010
xxxx (4/16) => 0011
xxxx (5/16) => 0100
xxxx (6/16) => 0101
xxxx (7/16) => 0110
xxxx (8/16) => 0111
xxxx (9/16) => 1000
xxxx (10/16) => 1001
xxxx (11/16) => 1010
xxxx (12/16) => 1011
xxxx (13/16) => 1100
xxxx (14/16) => 1101
xxxx (15/16) => 1110
xxxx (16/16) => 1111
0001 (1/1) => 0001
1010 (1/1) => 1010
0110 (1/1) => 0110

最好的问候。

【讨论】：

哇！天才:) 非常感谢！