如何计算位图？ [英] how to compute a bitmap?

问题描述

我要寻找一种方式来获得一个列表项目的所有组合。
我想是有两个渔政船阵，类似地以位图
例如位[] [] MYBITMAP;

I am looking for a way to get all combination of a list item. what i thinking is to have a two dimention array, similary to a bit map e.g bit[][] mybitmap;

例如，如果我有我的名单4项A，B，C，D
我希望我的位图是填充像这样

for example if i have 4 item in my list "A, B, C, D" i want my bitmap to be populate like this

A  B  C  D

0, 0, 0, 1  --> D
0, 0, 1, 0  --> C
0, 0, 1, 1  --> C, D
0, 1, 0, 0  --> B
0, 1, 0, 1
0, 1, 1, 0
0, 1, 1, 1
1, 0, 0, 0
1, 0, 0, 1
1, 0, 1, 0
1, 0, 1, 1  --> A, C, D
1, 1, 0, 0
1, 1, 0, 1
1, 1, 1, 0
1, 1, 1, 1  --> A, B, C, D

但我怎么能写一些C＃code来填充我的位图？
（PS：我的名单可能有大约80项到90，而不是100到200，刚刚确认）

but how can i write some C# code to populate my bit map? (PS: my list might have items around 80 to 90, not 100 to 200, just confirmed)

感谢

推荐答案

我相信你并不需要所有的组合存储在内存中。
从全零位（第一组合）阵列刚开始。为了得到下一个组合只需添加1〜previous组合的最后一位（这是很容易实现的操作）。等等。
低内存占用，支持高达2亿元的数字。 ：）

I believe you don't need to store all combinations in memory. Just start from array with all zero bits (first combination). To get next combination just add 1 to last bit of previous combination (it is easily implementing operation). And so on. Low memory usage, support of up to 2 billions of digits. :)

    private void button1_Click(object sender, EventArgs e)
    {
        string[] items = {"A", "B", "C", "D"};
        bool[] bits = new bool[items.Length];
        for (int i = 0; i < bits.Length; i++)
        {
            bits[i] = false;
        }
        while (!bits.All(x => x))
        {
            listBox1.Items.Add(string.Join(", ", GetCombination(items, bits)));
            AddBit(bits, bits.Length - 1);
        }
    }

    public string[] GetCombination(string[] items, bool[] bits)
    {
        List<string> combination = new List<string>();
        for (int i = 0; i < bits.Length; i++)
        {
            if (bits[i])
            {
                combination.Add(items[i]);
            }
        }
        return combination.ToArray();
    }

    private void AddBit(bool[] bits, int pos)
    {
        if (pos < 0)
        {
            // overflow :)
            return;
        }
        if (bits[pos])
        {
            bits[pos] = false;
            AddBit(bits, pos - 1);
        }
        else
        {
            bits[pos] = true;
        }
    }

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