算法问题，请高手解答

[zrcwzl]

请教各位高手，如何将以下数组合并？
Arr1 = {1,2,1};
Arr2 = {1,2,3};
Arr3 = {1,3,1};
Arr4 = {1,3,3};
Arr5 = {2,21};
Arr6 = {2,2,3};
Arr7 = {2,3,1};
Arr8 = {2,3,3};

按相同元素合并，得到最终数组
Arr = {1,2},{2,3},{1,3}

[ajianchina]

static void Main(string[] args)
{
	List<string[]> list = new List<string[]>();
	list.Add(new string[] { "1", "2", "1" });
	list.Add(new string[] { "1", "2", "3" });
	list.Add(new string[] { "1", "3", "1" });
	list.Add(new string[] { "1", "3", "3" });
	list.Add(new string[] { "2", "2", "1" });
	list.Add(new string[] { "2", "2", "3" });
	list.Add(new string[] { "2", "3", "1" });
	//list.Add(new string[] { "2", "3", "3" });
	//list.Add(new string[] { "3", "3", "3" });
	List<string[][]> list2 = list.Select(p => p.Select(s => new string[] { s }).ToArray()).ToList();
	string[][] arr;
	int count;
	while (true)
	{
		count = list2.Count;
		foreach (var ar in list2)
		{
			arr = list2.Where(p => p != ar && Eq(p, ar)).FirstOrDefault();
			if (arr == null) continue;
			for (int i = 0; i < ar.Length; i++)
			{
				arr[i] = ar[i].Union(arr[i]).ToArray();
			}
			list2.Remove(ar);
			break;//循环体被改变，立即跳出
		}
		if (list2.Count == count) break;
	}

	foreach(var x in list2)
	{
		Console.WriteLine(string.Join(",", x.Select(p=>string.Join("",p))));
	}
}
static bool Eq(string[][] a, string[][] b)
{
	int x = 0;
	for (int i = 0; i < a.Length; i++)
	{
		if (a[i].Length != b[i].Length || a[i].Intersect(b[i]).Count()!=a[i].Length) x++;
	}
	return x <= 1;
}

result: 1,23,13 2,2,13 2,3,1

[zrcwzl]

不知道Lambda的GroupBy方法能不能处理？

[娃都会打酱油了]

这什么意思啊，没看懂……

[bdmh]

你最终的结果是多维数组，没太清楚，结果和原始数据的对应

[zrcwzl]

改成字符串更好理解吧，具体合并顺序类似这样： Arr1 = {1,2,1}; Arr2 = {1,2,3}; 合并为 {1,2,13} Arr3 = {1,3,1}; Arr4 = {1,3,3}; 合并为{1,3,13} Arr5 = {2,2,1}; Arr6 = {2,2,3}; 合并为{2,2,13} Arr7 = {2,3,1}; Arr8 = {2,3,3}; 合并为{2,3,13} {1,2,13}和{1,3,13}合并为{1,23,13} {2,2,13}和{2,3,13}合并为{2,23,13} 最终合并为{12,23,13}

[zrcwzl]

合并到不能合并为止，每个数组长度相同，不仅限3位

[zrcwzl]

Arr5 = {2,21};这个我写错了，少了一个逗号，应该是Arr5 = {2,2,1};

[ajianchina]

List<string[]> list = new List<string[]>();
list.Add(new string[] { "1", "2", "1" });
list.Add(new string[] { "1", "2", "3" });
string[] arr = new string[list.First().Length];
for (int i = 0; i < arr.Length; i++)
{
	arr[i] = string.Join("", list.SelectMany(p=>p[i]).Distinct());
}

Console.WriteLine(string.Join(",", arr));
//result:1,2,13

[ajianchina]

没把你列的全放进去，放进去结果就是12,23,13

[zrcwzl]

谢谢ajianchina，不过有个问题，不是所有数组都能合并，如果再加个Arr9 = {3,3,3},合并后的结果应该是两个数组 {12,23,13}，{3,3,3}，因为{3,3,3}合并不了

[娃都会打酱油了]

static List<string[]> MergeArray(List<string[]> list)
{
    if (list == null || list.Count == 0)
    {
        throw new ArgumentException();
    }
    var x = (int)Math.Log(list.Count, 2);//保证只取最小整数部分
    if (x == 0)
    {
        if (list.Count == 1)
        {
            return list;
        }
        else
        {
            return new List<string[]> { list[0] };
        }
    }
    List<string[]> resultList = new List<string[]>();
    for (var i = 0; i < (int)Math.Pow(2, x); i += 2)
    {
        var li = list[i];
        var hi = list[i + 1];
        string[] tmp = new string[3];
        for (var j = 0; j < 3; j++)
        {
            tmp[j] = li[j];
            if (li[j] != hi[j])
            {
                tmp[j] += hi[j];
            }
        }
        resultList.Add(tmp);
    }
    return MergeArray(resultList);
}

严格按照你两两合并的顺序来，如果你将源数据改了，可以得到其他结果

List<string[]> list = new List<string[]>{
        new string[] {"1","2","1"},
new string[] {"1","2","3"},
new string[] {"1","3","1"},
new string[] {"1","3","3"},
new string[] {"2","2","1"},
new string[] {"2","2","3"},
new string[] {"2","3","1"},
new string[] {"2","3","3"}};

    var resultList = MergeArray(list);
    Console.WriteLine(string.Join(",", resultList[0]));

[zrcwzl]

如果去掉Arr8 = {2,3,3}，合并后最短的数组长度是3，分别是 {1,23,13}，{2,2,13}，{2,3,1}或者{12,2,13}，{1,3,13}，{2,3,1}

[zrcwzl]

你速度好快，厉害！

[zrcwzl]

谢谢starfd，不过结果不对

[娃都会打酱油了]

肯定不对啊，不是你要求的，我写代码时还没看到你那个要求 我的做法是过滤忽略你多出来的数据，因为没法合并，所以如果最小化修改方式的话，就是在调用外面时可以用循环来做

[娃都会打酱油了]

List<string[]> list = new List<string[]>{
        new string[] {"1","2","1"},
new string[] {"1","2","3"},
new string[] {"1","3","1"},
new string[] {"1","3","3"},
new string[] {"2","2","1"},
new string[] {"2","2","3"},
new string[] {"2","3","1"},
new string[] {"2","3","3"},
new string[] {"3","3","3"}};

    var resultList = new List<string[]>();
    var num = 0;
    while (list.Count != num)
    {
        var log = (int)Math.Log(list.Count - num, 2);
        var take = (int)Math.Pow(2, log);
        resultList.Add(MergeArray(list.Skip(num).Take(take).ToList())[0]);
        num += take;
    }
    Console.WriteLine(string.Join("   ", resultList.Select(x => string.Join(",", x))));

调用部分修正为这样就满足你的要求了

[同心圆的故事]

楼主是否需要考虑合并顺序？例如｛1，2，1｝&｛1，2，3｝与｛1，2，3｝&｛1，2，1｝的合并结果相同吗？

[凌晨4点5杀老大爷]

有没有顺序，没顺序的话 ，放到个数组里面 然后distinct 就可以了

[SunliyMonkey]

好抽象，楼主都不说下，合并规则

[zrcwzl]

谢谢ajianchina，你的结果是正确的