如果我有两个数组
A:[A,B]
B:[1,2,3]
如何创建像[A_1, A_2, A_3, B_1, B_2, B_3]
阵列的数量不规则,可能还有3个
A:[A,B]
B:[1,2,3]
C:[w,x,y,z]
D:[m,n]
E:[p,q,r]
我可以使用递归来解决它吗?
答案 0 :(得分:0)
这是对n-ary维度的简单迭代 - 不需要递归,只需要存储索引的数组。
static void Iterate(int[] iterators, ArrayList[] arrays) {
for (var j = iterators.Length - 1; j >= 0; j--) {
iterators[j]++;
if (iterators[j] == arrays[j].Count) {
if (j == 0) {
break;
}
iterators[j] = 0;
} else {
break;
}
}
}
static IList<string> Merge(ArrayList[] arrays) {
List<string> result = new List<string>();
int[] iterators = new int[arrays.Length];
while (iterators[0] != arrays[0].Count) {
var builder = new StringBuilder(20);
for(var index = 0; index < arrays.Length; index++) {
if (index > 0) {
builder.Append("_");
}
builder.Append(arrays[index][iterators[index]]);
}
result.Add(builder.ToString());
Iterate(iterators, arrays);
}
return result;
}
static void Main(string[] args) {
var list1 = new ArrayList();
var list2 = new ArrayList();
var list3 = new ArrayList();
list1.Add(1);
list1.Add(2);
list2.Add("a");
list2.Add("b");
list3.Add("x");
list3.Add("y");
list3.Add("z");
var result = Merge(new[] { list1, list2, list3 });
}
答案 1 :(得分:0)
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace arrconn {
class Program {
static string[] conn(params Array[] arrs) {
if(arrs.Length == 0) return new string[0];
if(arrs.Length == 1) {
string[] result = new string[arrs[0].Length];
for(int i = 0; i < result.Length; i++)
result[i] = arrs[0].GetValue(i).ToString();
return result; }
else {
string[] result = new string[arrs[0].Length*arrs[1].Length];
for(int i = 0; i < arrs[0].Length; i++)
for(int j = 0; j < arrs[1].Length; j++)
result[i*arrs[1].Length+j] = string.Format("{0}_{1}", arrs[0].GetValue(i), arrs[1].GetValue(j));
if(arrs.Length == 2) return result;
Array[] next = new Array[arrs.Length-1];
next[0] = result; Array.Copy(arrs, 2, next, 1, next.Length-1);
return conn(next);
}
}
static void Main(string[] args) {
foreach(string s in conn(
new string[] { "A", "B" },
new int[] { 1, 2, 3 },
new string[] { "x" },
new string[] { "$", "%", "#" }))
Console.WriteLine(s);
Console.Read();
}
}
}
答案 2 :(得分:0)
因此,我们定义了一个函数Merge
,它获取了stings列表的列表并将它们合并到你想要的可枚举字符串中
static void Main(string[] args)
{
var a = new[] { "A", "B" };
var b = new[] { "1", "2", "3" };
var c = new[] { "x", "y", "z", "w" };
var result = Merge(a, b, c);
foreach (var r in result)
{
Console.WriteLine(r);
}
}
public static IList<string> Merge(params IEnumerable<string>[] lists)
{
return Merge((IEnumerable<IEnumerable<string>>) lists);
}
public static IList<string> Merge(IEnumerable<IEnumerable<string>> lists)
{
var retval = new List<string>();
var first = lists.FirstOrDefault();
if (first != null)
{
var result = Merge(lists.Skip(1));
if (result.Count > 0)
{
foreach (var x in first)
{
retval.AddRange(result.Select(y => string.Format("{0}_{1}", x, y)));
}
}
else
{
retval.AddRange(first);
}
}
return retval;
}
如果您使用Lists
作为输入
public static IList<string> Merge(params IList<string>[] lists)
{
return Merge((IList<IList<string>>) lists);
}
public static IList<string> Merge(IList<IList<string>> lists, int offset = 0)
{
if (offset >= lists.Count)
return new List<string>();
var current = lists[offset];
if (offset + 1 == lists.Count) // last entry in lists
return current;
var retval = new List<string>();
var merged = Merge(lists, offset + 1);
foreach (var x in current)
{
retval.AddRange(merged.Select(y => string.Format("{0}_{1}", x, y)));
}
return retval;
}
答案 3 :(得分:0)
我猜您的输入是这样的:
var A = ["A","B"];
var B = [1,2,3];
var C = ["x","y","z","w"];
你想要获得的是:
var result = ["A_1_x", "A_1_y",...
"A_2_x", "A_2_y",...
"A_3_x", "A_3_y",...
"B_1_x", "B_1_y",...
...
..., "B_3_z", "B_3_w"];
我们将使用IEnumerable,因为它将简化我们的工作并让我们访问yield关键字。 首先,让我们来处理我们只收集两个集合的情况:
IEnumerable<string> ConcatEnumerables(IEnumerable<object> first, IEnumerable<object> second)
{
foreach (var x in first)
{
foreach (var y in second)
{
yield return x.ToString() + "_" + y.ToString();
}
}
}
然后我们可以递归地获取任意数量的集合:
IEnumerable<string> ConcatEnumerablesRec(IEnumerable<IEnumerable<object>> enums)
{
//base cases
if(!enums.Any())
{
return Enumerable.Empty<string>();
}
if (enums.Count() == 1)
{
return enums.First().Select(o => o.ToString());
}
//recursively solve the problem
return ConcatEnumerables(enums.First(), ConcatEnumerablesRec(enums.Skip(1));
}
现在你需要在结果上调用ToArray,如果你真的需要一个数组作为输出。
string[] Concatenator(params object[][] parameters)
{
return ConcatEnumerablesRec(parameters).ToArray();
}
答案 4 :(得分:0)
这应该可以解决问题。请注意,输入序列不必是数组 - 它们可以是实现IEnumerable<>
的任何类型。
另请注意,我们必须将值类型的序列置于<object>
的序列中,以便它们可分配给IEnumerable<object>
。
以下是可编辑的控制台应用演示代码:
using System;
using System.Collections.Generic;
using System.Linq;
namespace Demo
{
internal static class Program
{
static void Main()
{
string[] a = {"A", "B", "C", "D"};
var b = Enumerable.Range(1, 3); // <-- See how it doesn't need to be an array.
char[] c = {'X', 'Y', 'Z'};
double[] d = {-0.1, -0.2};
var sequences = new [] { a, b.Cast<object>(), c.Cast<object>(), d.Cast<object>() };
Console.WriteLine(string.Join("\n", Combine("", sequences)));
}
public static IEnumerable<string> Combine(string prefix, IEnumerable<IEnumerable<object>> sequences)
{
foreach (var item in sequences.First())
{
string current = (prefix == "") ? item.ToString() : prefix + "_" + item;
var remaining = sequences.Skip(1);
if (!remaining.Any())
{
yield return current;
}
else
{
foreach (var s in Combine(current, remaining))
yield return s;
}
}
}
}
}