数组求交集，并集

Question

我有n个数组，先求任意n-1个数组的交集
然后再求得到的新的n个数组的并集
例如：有5个数组
先求
a1 a2 a3 a4的交集b1
a1 a2 a3 a5的交集b2
a1 a2 a4 a5的交集b3
a1 a3 a4 a5的交集b4
a2 a3 a4 a5的交集b5
然后求得到的新的数组b1,b2,b3,b4,b5的并集
求方法

Answer 1

a1.Intersect(a2)  ==> a1和a2的交集a1.Union(a2)   ==>a1和a2的并集

Answer 2

List<int[]> list = new List<int[]>();
            for (int i = 0; i < 5; i++)
                list.Add(Enumerable.Range(1, 20).OrderBy(T => Guid.NewGuid()).Take(10).ToArray());
            list.ForEach(t => Console.WriteLine(string.Join(",", t.Select(T => T.ToString()).ToArray())));
            int[] source = new int[0];
            for (int i = 0; i < 5; i++)
            {
                List<int[]> lst = new List<int[]>();
                list.Select((t, index) => { if (index != i)lst.Add(t); return 0; }).ToArray();
                source = source.Union(lst.Aggregate((a, b) => a.Intersect(b).ToArray())).ToArray();
            }
            Console.WriteLine(string.Join(",", source.Select(T => T.ToString()).ToArray()));

Answer 3

    List<int[]> al = new List<int[]>();
            al.Add(new int[] { 1 });
            al.Add(new int[] { 1, 2, 3, 4 });
            al.Add(new int[] { 1,3 });
            al.Add(new int[] { 3, 4 });
            al.Add(new int[] { 1, 2 });            List<int[]> al2 = new List<int[]>();
            for (int i = 0; i < al.Count; i++)
            {
                int[] s = null;
                for (int j = 0; j < al.Count; j++)
                {
                    if (i != j)
                    {
                        if (j == 0)
                        {
                            s = al[j];
                        }
                        else
                        {
                            if (i == 0) 
                            {
                                if (j == 1)
                                {
                                    s = al[j];
                                }
                                else 
                                {
                                    s = (s.Intersect(al[j])).ToArray();
                                }
                            }
                            else
                            {
                                s = (s.Intersect(al[j])).ToArray();
                            }
                        }
                    }
                }
                al2.Add(s);
            }            int[] t = null;
            for (int k = 0; k < al2.Count; k++)
            {
                if (k == 0)
                {
                    t = al2[0];
                }
                else
                {
                    t = (t.Intersect(al2[k])).ToArray();
                }
            }            //最后得到的t就是你所要的数据

Answer 4

方法我自己想到了
有n个数组，先求任意n-1个数组的交集
然后再求得到的新的n个数组的并集
说明里边的元素至少要出现n-1次
所以我的做法就是先把所有的n个数组求并集
然后提取出现次数大于等于n-1次的元素
然后去重还望大家指点