3秒搞定！~~ 一亿数据获取前100个最大值

整合网络上的算法。根据我的思路。计算一亿个数字中最大的前100个值。

昨晚效率还是很低。今天搞的算法。只需要3秒钟。获取前100个前1000个速度都非常快。

算法原理：

把一亿个数字的前100个首先放入数组。然后把最小值放在ary[0]。

然后再，循环100到一亿之间的。每次循环判断当前数字是否大于ary[0]

当大于时，当前数字放入ary[0] 并再次重构数组最小值进入ary[0] 以此类推。

当循环完这一亿个数字后。最大的前100个数字就出来了。

源码分享地址：http://download.csdn.net/download/yjflinchong/4275241

腾讯面试题：一亿数字获取前100个最大的数字办法

比较笨的办法。效率有点低。只是实现了功能。期待牛人的算法。

我弄了个最佳方案 http://blog.csdn.net/yjflinchong/article/details/7533972 3秒就搞定了

一亿数字获取前100个最大的数字这个方案需要700秒

///http://blog.csdn.net/yjflinchong/article/details/7532018

package com.my.util;

import java.io.*;
import java.util.*;
import java.net.*;

public class WebTest {

public static int last = 333333333;
   public static int max = 100000000;//数据总数
   public static int sp = 1000000;//分割数据条数
   public static int[] ary = new int[100];


   public static void main(String[] args) {
       splitFile();
       Date d1 = new Date();
       find("d:/file/file.txt");
       System.out.println(Arrays.toString(ary)+"========");
       Date d2 = new Date();
       System.out.println(d2.getTime()-d1.getTime());
   }


   public static void splitFile(){
       StringBuffer str = new StringBuffer("");
       int num = 1;
       for (int i = 1; i < (max+1); i++) {
           str.append(i+"\t\n");
           if(i%1000000==0){
               appendToFile("d:/file/file.txt", str.toString());
               str = new StringBuffer("");
               num++;
           }
       }
       appendToFile("d:/file/file.txt", last+"");
   }

public static void appendToFile(String file,String text){
       try
       {
           FileWriter fw =new FileWriter(file,true);
           BufferedWriter bw=new BufferedWriter(fw);
       bw.write(text);
       bw.flush();
       }catch (Exception e) {
       }
       }
       public static String readFile(String path){
           StringBuffer str = new StringBuffer("");
           try {
                String line = null;

                BufferedReader reader = new BufferedReader(new FileReader(path));
                while ((line = reader.readLine()) != null) {
                   str.append(line);
                }
                reader.close();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str.toString();
       }
       public static void find(int[] bak){
           for (int i = 0; i < bak.length; i++) {
               ary[0] = bak[i];
               sort(ary);
           }
       }///http://blog.csdn.net/yjflinchong/article/details/7532018
       public static void find(String path){
           try {///http://blog.csdn.net/yjflinchong/
                String line = null;
                BufferedReader reader = new BufferedReader(new FileReader(path));
                while ((line = reader.readLine()) != null) {
                   ary[0] = Integer.parseInt(line.trim());
                   sort(ary);
                }
                reader.close();
            } catch (Exception e) {
                e.printStackTrace();
            }
       }
       ///http://blog.csdn.net/yjflinchong/article/details/7532018
       public static void sort(int[] array){
            for(int i = 0; i < array.length - 1; i++){
                //当前值当作最小值
                int min = array[i];
                for(int j = i+1; j < array.length; j++){
                    if(min>array[j]){
                        //如果后面有比min值还小的就交换
                        min = array[j];
                        array[j] = array[i];
                        array[i] = min;
                    }
                }
            }
        }

}

一亿个数字判断其中相同数字的办法

package com.my.util;

//http://blog.csdn.net/yjflinchong
public class Test {

   int fnum = 21000000;

   public static void main(String[] args) {
       Test t = new Test();
       t.find();
   }


   public void find() {
       int total = 100000000;
       int size = total%32==0?total/32:total/32+1;
       int [] mBits=new int[size];
       for(int i=0;i<total;i++) {
           int num = getNum(i);
           if(get(mBits,num)) {
               System.out.println(num);
               break;
           }
          set1(mBits,num);
       }
   }
   //http://blog.csdn.net/yjflinchong
   public int getNum(int i) {
       //设定模拟重复的那个数字 fnum
       if(i==(fnum+1)){
           i--;
       }
       return i;
   }
   public void set1(int [] mBits, int pos) {
       int index = ( int )Math.floor(pos/32f);
       mBits[index] = mBits[index] | (1 <<(31-pos%32 ));
   }
   public boolean get(int [] mBits, int pos){
       int index = ( int )Math.floor(pos/32f);
       return mBits[index] == (mBits[index] | 1 <<(31-pos% 32 ));
   }

}