算法导论83(a)习题解答(基数排序)

CLRS 8-3(a) : 给定一个整数数组，其中不同的整数中包含的数字个数可能不同，但该数组中，所有整数中总的数字数为n。说明如何在O(n)时间内对该数组进行排序。

算法思想：

1)先把各个数字按位数分组。

2)然后对各个分数进行基数排序

3)将各个组连接起来即可

这个题目基本上重复了算法导论8.3-4习题解答(基数排序) 的核心算法.

#include <iostream>
#include <fstream>
using namespace std;
void radix_sort(int* a, constint d, constint length, int radix);
int getBit(int m, int i, int radix);
int pow2(int a, int b);
int main()
{
      const int LEN =50000;
      int a[LEN];
      for(int i =0; i < LEN; i++)
  {
    a[i] = rand()%100000;
  }
      //一位数、二位数、三位数、四位数的个数
  int one = 0; 
      int ten = 0; 
      int hundred = 0; 
      int thousand = 0; 
      int ten_thousand = 0;
      //保存一位数、二位数、三位数、四位数相应的数组
  int* one_arr = new int[LEN];
      int* ten_arr =new int[LEN];
      int* hundred_arr =new int[LEN];
      int* thounsand_arr =new int[LEN];
      int* ten_thousand_arr =new int[LEN];
      for(int i =0; i < LEN; i++)
  {
            if(a[i] >=0&& a[i] <10)
      one_arr[one++] = a[i];
            else if(a[i] >=10&& a[i] <100)
      ten_arr[ten++] = a[i];
            else if(a[i] >=100&& a[i] <1000)
      hundred_arr[hundred++] = a[i];
            else if(a[i] >=1000&& a[i] <10000)
      thounsand_arr[thousand++] = a[i];
            else
      ten_thousand_arr[ten_thousand++] = a[i];
  }
    
  radix_sort(one_arr, 1, one, 10);
  radix_sort(ten_arr, 2, ten, 10);
  radix_sort(hundred_arr, 3, hundred, 10);
  radix_sort(thounsand_arr, 4, thousand, 10);
  radix_sort(ten_thousand_arr, 5, ten_thousand, 10);
      int one_to_ten = one + ten;
      int one_to_hundred = one + ten + hundred;
      int one_to_thousand = one + ten + hundred + thousand;
      int one_to_ten_thousand = one + ten + hundred + thousand + ten_thousand;
      for(int i =0; i < LEN; i++)
  {
            if(i < one)
      a[i] = one_arr[i];
            else if(i < one_to_ten)
      a[i] = ten_arr[i - one];
            else if(i < one_to_hundred)
      a[i] = hundred_arr[i - one_to_ten];
            else if(i < one_to_thousand)
      a[i] = thounsand_arr[i - one_to_hundred];
            else if(i < one_to_ten_thousand)
      a[i] = ten_thousand_arr[i - one_to_thousand];
  }
      const char* file_name = "8-3(a).txt";
  ofstream out_to_file;
  out_to_file.open(file_name);
      for(int i =0; i < LEN; i++)
    out_to_file<<a[i]<<endl;
  out_to_file.close();
  delete one_arr;
  delete ten_arr;
  delete hundred_arr;
  delete thounsand_arr;
  delete ten_thousand_arr;
      return 0;
}
void radix_sort(int* a, constint d, constint length, int radix)
{
      //存储a中每个元素的余数
  int* remainder =new int[length];
      //统计余数(等同于计数排序中的c)
  int* c =new int[radix];
      //保存排序之后的a
  int* b =new int[length];
      for(int i =0; i < radix; i++)
    c[i] =0;
      for(int i =0; i < d; i++)
  {
            for(int j =0; j < length; j++)
    {
                  int temp = getBit(a[j], i, radix);
      remainder[j] = temp;
      c[temp]++;
    }
            for(int k =1; k < radix; k++)
    c[k] = c[k] + c[k -1];
            for(int k = length -1; k >=0; k--)
    {
      b[--c[remainder[k]]] = a[k];
                  //c[remainder[k]]--;
    }
            for(int k =0; k < radix; k++)
      c[k] =0;
            for(int i =0; i < length; i++)
      a[i] = b[i];
  }
  delete remainder;
  delete c;
  delete b;
}
//得到相应位上的数值
int getBit(int m, int i, int radix)
{
       return (m%(pow2(radix, i +1)))/pow2(radix, i);
}
//只处理指数b>=0
int pow2(int a, int b)
{
      if(b >0)
  {
            int result = a;
            for(int i =0; i < b -1; i++)
      result = result*a;
            return result;
  }
      else if(b ==0)
            return 1;
      else
  {
    cout<<"不处理指数小于零的情况"<<endl;
            return -1;
  }
}