用C#实现全组合!!!
发布网友
发布时间：2024-10-08 09:34
共2个回答
热心网友
时间：2024-10-25 16:54
//参考链接：http://www.cnblogs.com/snowdust/archive/2010/01/20/1652161.html
class Program
{
static void Main(string[] args)
{
int[] array = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 };
//求排列
List<int[]> lst_Permutation = PermutationAndCombination<int>.GetPermutation(array, 5);
//求组合
List<int[]> lst_Combination = PermutationAndCombination<int>.GetCombination(array, 5);
Console.ReadLine();
}
}
public class PermutationAndCombination<T>
{
/// <summary>
/// 交换两个变量
/// </summary>
/// <param name="a">变量1</param>
/// <param name="b">变量2</param>
public static void Swap(ref T a, ref T b)
{
T temp = a;
a = b;
b = temp;
}
/// <summary>
/// 递归算法求数组的组合(私有成员)
/// </summary>
/// <param name="list">返回的范型</param>
/// <param name="t">所求数组</param>
/// <param name="n">辅助变量</param>
/// <param name="m">辅助变量</param>
/// <param name="b">辅助数组</param>
/// <param name="M">辅助变量M</param>
private static void GetCombination(ref List<T[]> list, T[] t, int n, int m, int[] b, int M)
{
for (int i = n; i >= m; i--)
{
b[m - 1] = i - 1;
if (m > 1)
{
GetCombination(ref list, t, i - 1, m - 1, b, M);
}
else
{
if (list == null)
{
list = new List<T[]>();
}
T[] temp = new T[M];
for (int j = 0; j < b.Length; j++)
{
temp[j] = t[b[j]];
}
list.Add(temp);
}
}
}
/// <summary>
/// 递归算法求排列(私有成员)
/// </summary>
/// <param name="list">返回的列表</param>
/// <param name="t">所求数组</param>
/// <param name="startIndex">起始标号</param>
/// <param name="endIndex">结束标号</param>
private static void GetPermutation(ref List<T[]> list, T[] t, int startIndex, int endIndex)
{
if (startIndex == endIndex)
{
if (list == null)
{
list = new List<T[]>();
}
T[] temp = new T[t.Length];
t.CopyTo(temp, 0);
list.Add(temp);
}
else
{
for (int i = startIndex; i <= endIndex; i++)
{
Swap(ref t[startIndex], ref t[i]);
GetPermutation(ref list, t, startIndex + 1, endIndex);
Swap(ref t[startIndex], ref t[i]);
}
}
}
/// <summary>
/// 求从起始标号到结束标号的排列，其余元素不变
/// </summary>
/// <param name="t">所求数组</param>
/// <param name="startIndex">起始标号</param>
/// <param name="endIndex">结束标号</param>
/// <returns>从起始标号到结束标号排列的范型</returns>
public static List<T[]> GetPermutation(T[] t, int startIndex, int endIndex)
{
if (startIndex < 0 || endIndex > t.Length - 1)
{
return null;
}
List<T[]> list = new List<T[]>();
GetPermutation(ref list, t, startIndex, endIndex);
return list;
}
/// <summary>
/// 返回数组所有元素的全排列
/// </summary>
/// <param name="t">所求数组</param>
/// <returns>全排列的范型</returns>
public static List<T[]> GetPermutation(T[] t)
{
return GetPermutation(t, 0, t.Length - 1);
}
/// <summary>
/// 求数组中n个元素的排列
/// </summary>
/// <param name="t">所求数组</param>
/// <param name="n">元素个数</param>
/// <returns>数组中n个元素的排列</returns>
public static List<T[]> GetPermutation(T[] t, int n)
{
if (n > t.Length)
{
return null;
}
List<T[]> list = new List<T[]>();
List<T[]> c = GetCombination(t, n);
for (int i = 0; i < c.Count; i++)
{
List<T[]> l = new List<T[]>();
GetPermutation(ref l, c[i], 0, n - 1);
list.AddRange(l);
}
return list;
}
/// <summary>
/// 求数组中n个元素的组合
/// </summary>
/// <param name="t">所求数组</param>
/// <param name="n">元素个数</param>
/// <returns>数组中n个元素的组合的范型</returns>
public static List<T[]> GetCombination(T[] t, int n)
{
if (t.Length < n)
{
return null;
}
int[] temp = new int[n];
List<T[]> list = new List<T[]>();
GetCombination(ref list, t, t.Length, n, temp, n);
return list;
}
}
热心网友
时间：2024-10-25 16:56
{
List<List<int>> result = new List<List<int>>(); //保存结果的集合；
for(int i=0;i<arr.Length;i++ )
{
List<int> myList = arr.ToList<int>();//将数组转化为List
List<int> arrlist = new List<int>();
for(int j=1;j<arr.Length;j++)
{
if(myList.Count-i>=2)//保证myList中元素至少有2个；
{
arrlist=myList.GetRange(i,2);//从myList中获取从索引i开始长度为2的子集
if(!result.Contains(arrlist))//确保没有重复
{
result.Add(arrlist);//将结果存入result
}
}
if (i < myList.Count-1)
{
myList.RemoveAt(i +1);
}
}
}
return result;//返回结果
}
该函数其实就是遍历每一个数组元素，将该包含该元素长度为2的子集装入arrlist..
之后只需要用result里的元素替换掉原数组相同的数字然后迭代。
