自己在一个搜索程序中遇到了这样一个问题:怎么从数组(集合)中返回指定长度的子数组(集合)。比如数组{1,2,3,4},现在要返回所有长度为n=2的子数组,即{1,2}{1,3}{1,4}{2,3}{2,4}{3,4}。如果这个n在写代码时就确定,那就用n层循环可以很简单的实现。但是,关键在于n是在程序运行时才知道的,这样就不能只能用循环了。
想了几天后,才完全实现了这个功能。
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Code
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using System;
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using System.Collections.Generic;
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using System.Linq;
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using System.Linq.Expressions;
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using System.Text;
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using System.Runtime.InteropServices;
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using System.Drawing;
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using System.Diagnostics;
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using System.Threading;
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using System.IO;
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using System.Runtime.Serialization;
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using System.Runtime.Serialization.Formatters.Binary;
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namespace ConsoleApplication1
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{
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class Program
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{
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static void Main(string[] args)
{
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//要操作的集合
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List<int> array = new List<int>();
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for (int i = 1; i <= 8; i++)
{
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array.Add(i);
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}
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//结果集合
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List<List<int>> result = Cal(array, 2);
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for (int i = 0; i < result.Count; i++)
{
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for (int j = 0; j < result[i].Count; j++)
{
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Console.Write(result[i][j] + " ");
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}
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Console.WriteLine();
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}
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Console.WriteLine(result.Count);
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Console.ReadKey();
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}
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//
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static List<List<int>> Cal(List<int> array, int n)
{
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List<List<int>> result = new List<List<int>>();
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List<int> one = new List<int>();
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for (int i = 0; i < array.Count; i++)
{
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one.Add(array[i]);
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Add(array.GetRange(i + 1, array.Count - 1 - i), n - 1, result, one);
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if (one.Count != 0)
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one.RemoveAt(one.Count - 1);
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//这里很可以用one.Clear();
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//循环完一次后,就可以清除one,然后再重新开始
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}
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return result;
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}
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//用于递归的函数
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static void Add(List<int> array, int n, List<List<int>> result, List<int> one)
{
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//如果n=0,就表明one的Count属性已等于n
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if (n == 0)
{
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result.Add(Clone(one) as List<int>);
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//进行下一次之前,移除最后一个元素
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one.RemoveAt(one.Count - 1);
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}
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else
{
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for (int i = 0; i < array.Count; i++)
{
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one.Add(array[i]);
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//在这里进行递归,同时n-1
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Add(array.GetRange(i + 1, array.Count - 1 - i), n - 1, result, one);
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}
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//一轮结束后,移除最后一个元素
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one.RemoveAt(one.Count - 1);
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}
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}
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//用来复制对象的函数
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public static object Clone(object obj)
{
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using (MemoryStream ms = new MemoryStream())
{
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IFormatter formattor = new BinaryFormatter();
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formattor.Serialize(ms, obj);
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ms.Seek(0, SeekOrigin.Begin);
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return formattor.Deserialize(ms);
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}
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}
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}
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}
上面的算法,用到了递归,关键在于设置了一个临时变量List<int> one,每递归一个就在one中添加一个元素,直到one的Count==n。
这是自己的想法,这个题目应该是很简单的,不知道有没有高手给出更简单的答案!!