我有一个字符串“ABCDE” 的ArrayList我想的方法与给定的ArrayList的所有可能的组合,以返回另一个数组列表:在C#(如AB,AC,AD ...)如何获得一个数组列表的所有组合?
有谁知道一个简单的方法?
NB:长度为2的所有可能的组合,如果长度是可变的,效果会更好(可以更改)
我有一个字符串“ABCDE” 的ArrayList我想的方法与给定的ArrayList的所有可能的组合,以返回另一个数组列表:在C#(如AB,AC,AD ...)如何获得一个数组列表的所有组合?
有谁知道一个简单的方法?
NB:长度为2的所有可能的组合,如果长度是可变的,效果会更好(可以更改)
修饰您的评论需要长度为两个的组合:应对
string s = "abcde";
var combinations = from c in s
from d in s.Remove(s.IndexOf(c), 1)
select new string(new[] { c, d });
foreach (var combination in combinations) {
Console.WriteLine(combination);
}
您编辑任何长度:
static IEnumerable<string> GetCombinations(string s, int length) {
Guard.Against<ArgumentNullException>(s == null);
if (length > s.Length || length == 0) {
return new[] { String.Empty };
if (length == 1) {
return s.Select(c => new string(new[] { c }));
}
return from c in s
from combination in GetCombinations(
s.Remove(s.IndexOf(c), 1),
length - 1
)
select c + combination;
}
用法:
string s = "abcde";
var combinations = GetCombinations(s, 3);
Console.WriteLine(String.Join(", ", combinations));
输出:
abc, abd, abe, acb, acd, ace, adb, adc, ade, aeb, aec, aed, bac, bad, bae, bca,
bcd, bce, bda, bdc, bde, bea, bec, bed, cab, cad, cae, cba, cbd, cbe, cda, cdb,
cde, cea, ceb, ced, dab, dac, dae, dba, dbc, dbe, dca, dcb, dce, dea, deb, dec,
eab, eac, ead, eba, ebc, ebd, eca, ecb, ecd, eda, edb, edc
你也应该做var retComb = GetCombinations(s,2).Distinct()。其中(p => p [0]> = p [p.Length]);删除多余的项目,以及项目的反向,例如'aabd'测试你的方法 – 2010-11-14 10:25:42
这是我能返回T类型的所有组合的通用功能:
static IEnumerable<IEnumerable<T>> GetCombinations<T>(IEnumerable<T> list, int length)
{
if (length == 1) return list.Select(t => new T[] { t });
return GetCombinations(list, length - 1)
.SelectMany(t => list, (t1, t2) => t1.Concat(new T[] { t2 }));
}
用法:
Console.WriteLine(
string.Join(", ",
GetCombinations("abcde".ToCharArray(), 2).Select(list => string.Join("", list))
)
);
输出:
aa, ab, ac, ad, ae, ba, bb, bc, bd, be, ca, cb, cc, cd, ce, da, db, dc, dd, de, ea, eb, ec, ed, ee
已更新 对于其他情况,请参阅我的回答here。排列和k-组合等中数字阵列
这是重复而不是组合的置换。 http://www.mathsisfun.com/combinatorics/combinations-permutations.html – 2014-02-26 08:33:57
@HalilKaskavalci,请查看[我的另一个答案](http://stackoverflow.com/a/10629938/1251423)了解更多选项。 – Pengyang 2014-03-13 07:36:29
现在看起来好多了。 – 2014-03-13 12:35:20
组合通过仅使用阵列和递归:
static int n = 4;
int[] baseArr = { 1, 2, 3, 4 };
int[] LockNums;
static void Main(string[] args)
{
int len = baseArr.Length;
LockNums = new int[n];
for (int i = 0; i < n; i++)
{
int num = baseArr[i];
DoCombinations(num, baseArr, len);
//for more than 4 numbers the print screen is too long if we need to check the result next line will help
//Console.ReadLine();
}
}
private void DoCombinations(int lockNum, int[] arr, int arrLen)
{
int n1 = arr.Length;
// next line shows the difference in length between the previous and its previous array
int point = arrLen - n1;
LockNums[n - arr.Length] = lockNum;
int[] tempArr = new int[arr.Length - 1];
FillTempArr(lockNum, arr, tempArr);
//next condition will print the last number from the current combination
if (arr.Length == 1)
{
Console.Write(" {0}", lockNum);
Console.WriteLine();
}
for (int i = 0; i < tempArr.Length; i++)
{
if ((point == 1) && (i != 0))
{
//without this code the program will fail to print the leading number of the next combination
//and 'point' is the exact moment when this code has to be executed
PrintFirstNums(baseArr.Length - n1);
}
Console.Write(" {0}", lockNum);
int num1 = tempArr[i];
DoCombinations(num1, tempArr, n1);
}
}
private void PrintFirstNums(int missNums)
{
for (int i = 0; i < missNums; i++)
{
Console.Write(" {0}", LockNums[i]);
}
}
private void FillTempArr(int lockN, int[] arr, int[] tempArr)
{
int idx = 0;
foreach (int number in arr)
{
if (number != lockN)
{
tempArr[idx++] = number;
}
}
}
private void PrintResult(int[] arr)
{
foreach (int num in arr)
{
Console.Write(" {0}", num);
}
}
你好,这是我对数字组合的解决方案。该解决方案只使用数组。我正在做这篇文章,因为我没有遇到任何地方的数组解决方案。代码设置为快速演示,否则它应该适用于任何数组长度。 – 2017-10-29 10:55:10
你指的是长度= 2的所有组合? – 2010-11-13 21:55:11
是的,遗憾忘记提到 – Mona 2010-11-13 21:55:48
所有可能的长度组合2 – Mona 2010-11-13 21:56:39