文件记录排序

时间：2009-08-09 来源：ubuntuer

你现在有一个文件，文件中顺序存有N个记录，R1,R2,...,RN,这些记录不是有序的，但是你知道一个整数M,这些记录满足R1<R2<...<RM以及RM+1<RM+2<...RN.

1,设计一个算法或编写一个程序，将文件中的记录排序为R1'<R2',<…<,RN',算法或程序读取文件的次数为O(N),不限内存使用，

2,设计一个算法或编写一个程序，将文件中的记录排序为R1'<R2'<...<RN',算法或程序读写文件的次数为O(N)，空间复杂度为O(1)，(亦即，你使用的内存大小和M,N均无关。) 1.就是两个有序数组的合并了,时间复杂度确实是O(N)，好说 2.第一个限制空间, or 内存不够时就是外部排序了.主要就是读写文件了.

#include <stdio.h>
#include <stdlib.h>

#define MAX_LEN 10
#define M 5
#define N 10

int process_file1(int A[])
{
   FILE* fp1 = fopen("record_bak.txt","w");
   int i = 0;
   int j = M;
   char line[MAX_LEN];

   while(i<M && j<N)
   {
     if(A[i]<A[j])
      {
        sprintf(line, "%d\n", A[i]);
        fwrite(line, 1, strlen(line), fp1);
        i++;
      }
     else
      {
        sprintf(line, "%d\n", A[j]);
        fwrite(line, 1, strlen(line), fp1);
        j++;
      }
   }

   if(i==M)
    {
     while(j<N)
     {
      sprintf(line, "%d\n", A[j]);
      fwrite(line, 1, strlen(line), fp1);
      j++;
     }
    }
   else if(j==N)
    {
     while(i<M)
     {
      sprintf(line, "%d\n", A[i]);
      fwrite(line, 1, strlen(line), fp1);
      i++;
     }
    }
fclose(fp1);
}

int process_file2()
{
  FILE* fp = NULL;
  FILE* fp1 = NULL;
  FILE* fp2 = NULL;

  char line1[MAX_LEN];
  char line2[MAX_LEN];

  int num1 = 0;
  int num2 = 0;
  int flag = 0;

  int i = 0;

  fp = fopen("record.txt","r");
  fp1 = fopen("record_tmp1.txt","w");
  fp2 = fopen("record_tmp2.txt","w");

  while(fgets(line1, MAX_LEN, fp))
   {
     printf("line is %s",line1);
     if(i<M)
       {
         printf("%d\n",fwrite(line1, 1, strlen(line1), fp1));
         i++;
       }
     else
       {
         printf("%d\n",fwrite(line1, 1, strlen(line1), fp2));
         i++;
       }
   }
   fclose(fp);
   fclose(fp1);
   fclose(fp2);

   fp = fopen("record.txt","w");
   fp1 = fopen("record_tmp1.txt","r");
   fp2 = fopen("record_tmp2.txt","r");

   fgets(line1, MAX_LEN, fp1);
   fgets(line2, MAX_LEN, fp2);
   while(1)
   {
      num1 = atoi(line1);
      num2 = atoi(line2);
      if(num1 < num2)
       {
         fwrite(line1, 1, strlen(line1), fp);
         if(!fgets(line1, MAX_LEN, fp1))
          {
           fwrite(line2, 1, strlen(line1), fp);
           flag = 1;
           break;
          }
       }
      else
       {
         fwrite(line2, 1, strlen(line2), fp);
         if(!fgets(line2, MAX_LEN, fp2))
          {
           fwrite(line1, 1, strlen(line1), fp);
           flag = 2;
           break;
          }
       }
    }
   if(flag == 1)
   {
     while(fgets(line2, MAX_LEN, fp2))
       fwrite(line2, 1, strlen(line2), fp);
   }
   else if(flag == 2)
   {
     while(fgets(line1, MAX_LEN, fp1))
       fwrite(line1, 1, strlen(line2), fp);
   }
   fclose(fp);
   fclose(fp1);
   fclose(fp2);
   remove("record_tmp1.txt");
   remove("record_tmp2.txt");
}

int main(int argc, char *argv[])
{
  int A[N];
  int i = 0;
  FILE* fp = NULL;
  char line[MAX_LEN];

  fp = fopen("record.txt","r");
  while(fgets(line, MAX_LEN, fp))
   {
     A[i] = atoi(line);
     i++;
   }
  fclose(fp);
  process_file1(A);
  process_file2(A);
  system("PAUSE");
  return 0;
}