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  • 从十亿个数中找出前1000个最大的数的算法

        10亿个数中找出1000个最大的数的算法思路:

       1,先拿出前1000个数字,并排序,找出最小值为 minValue .
       2,  然后再依次拿出1000个数字,找出最大值为 tempMaxValue.
       3,  如果tempMaxValue 比 minValue 大, 则将 theMaxValue 放入前1000个数中,再排序并找出minValue .
       4 .. 依此类推。 即可得到1000个最大的数。 


       以下是本人测试的代码 。可以参考,欢迎高手指正,谢谢!

    using System;
    using System.Collections.Generic;
    using System.Linq;
    using System.Text;
    using System.Collections;

    namespace test
    {
        public class Class1
        {
            private  const int MaxLength = 100000000;
            private const int EachCount = 1000;

            public static void TestDirectSort()
            {
                Console.WriteLine(" test start :" + DateTime.Now.ToLongTimeString());

                int[] orignList = new int[MaxLength];

                Random rd = new Random();
                for (int i = 0; i < MaxLength; i++)
                {
                    orignList[i] = rd.Next();
                }

                ArrayList list = new ArrayList(MaxLength);
                list.AddRange(orignList);
                list.Sort();

                int[] retList = new int[1000];
                int j = 0;
                for (int i = MaxLength - 1; i > MaxLength - 1000; i++)
                {
                    retList[j] = orignList[i];
                    j++;
                }

                Console.WriteLine(" test end : " + DateTime.Now.ToLongTimeString());

                foreach (int i in list)
                {
                    Console.WriteLine(i.ToString());
                }
                Console.WriteLine("end: " + DateTime.Now.ToLongTimeString());

            }

            public static void TestSort()
            {
                Console.WriteLine(" test start :"  + DateTime.Now.ToLongTimeString());

                int[] orignList = new int[MaxLength] ;
               
                Random rd = new Random();
                for (int i = 0; i < MaxLength; i++)
                {
                    orignList[i] = rd.Next();
                }

                ArrayList foreList = new ArrayList();
                if (MaxLength > EachCount)
                {
                    for (int i = 0; i < EachCount; i++)
                    {
                        foreList.Add(orignList[i]);
                    }

                    foreList.Sort();
                    int addValue = EachCount;
                    for (int j = 1; j < MaxLength / EachCount +1 ; j++)
                    {
                        ArrayList tempList = new ArrayList();
                        for (int k = j * addValue; k < j * addValue + EachCount; k++)
                        {
                            if (k >= MaxLength) break;
                            //tempList[k - j * addValue] = orignList[k];
                            tempList.Add(orignList[k]);
                        }

                        tempList.Sort();
                        for (int m = tempList.Count -1; m >=0 ; m--)
                        {
                            if ((int)tempList[m] <= (int)foreList[0])
                                break;
                           
                            foreList[0] = tempList[m];
                            foreList.Sort();
                        }
                    }
                }
                else
                {
                    foreList.AddRange(orignList);
                    foreList.Sort();
                }

                Console.WriteLine(" test end : " + DateTime.Now.ToLongTimeString());

                foreach (int i in foreList)
                {
                    Console.WriteLine(i.ToString());
                }
                Console.WriteLine("end: " + DateTime.Now.ToLongTimeString());
            }
        }

        public class client
        {
            public static void Main()
            {
                Class1.TestSort();

               // Class1.TestDirectSort();

                Console.Read();

            }
        }
    }

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  • 原文地址:https://www.cnblogs.com/relang99/p/1202073.html
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