一道水题

 
 

  
  题目：
 
 
 
 

  
  

   
   Common courses
  
  
  
  

  
  

   
   
A teacher wants to compare the performance of two students. To understand them better, he’s looking at all the other courses they took, but it’s hard to spot the common courses just from a glance.
   
   
Given two arrays that contain the course IDs of two different students
   
   
Your task is to
   
   
write a function that prints to standard output (stdout) all the course IDs that are contained in both arrays, sorted in ascending order, one per line
   
   
Note that your function will receive the following arguments:
   
   
courses1
     
     
which is the list of course IDs for the first student
courses2
     
     
which is the list of course ids for the second student
   
   
Data constraints
   
   
the length of the array given as input will not exceed 1000 elements
   
   
Example
   
   
Input
Output
courses1: 1, 2, 8, 4, 5, 8, 3 courses2: 8, 2, 2, 7, 10
2 8
  
  

 
 
 
 


 
 
 
 

  
  程序：
 
 
 
 

  
  

 
 
 
 

  
  

 
 
 
 

  
  #include <stdio.h>
 
 
 
 

  
  #include <stdlib.h>
 
 
 
 


 
 
 
 

  
  int 
  
  compare
  
  (
  
  const 
  
  void 
  
  *
  
  a
  
  ,
  
  const 
  
  void 
  
  *
  
  b
  
  ) 
  
  {
 
 
 
 
    
  
  return 
  
  *
  
  (
  
  int 
  
  *
  
  )
  
  a 
  
  - 
  
  *
  
  (
  
  int 
  
  *
  
  )
  
  b
  
  ;
 
 
 
 

  
  }
 
 
 
 


 
 
 
 

  
  void 
  
  get_common_courses
  
  (
  
  int 
  
  *
  
  courses1
  
  , 
  
  int 
  
  courses1_length
  
  , 
  
  int 
  
  *
  
  courses2
  
  , 
  
  int 
  
  courses2_length
  
  ) 
  
  {
 
 
 
 
    
  
  int 
  
  i
  
  ,
  
  j
  
  ;
 
 
 
 


 
 
 
 
  
  
  qsort
  
  (
  
  courses1
  
  ,
  
  courses1_length
  
  ,
  
  sizeof
  
  (
  
  int
  
  ),
  
  compare
  
  );
 
 
 
 
  
  
  qsort
  
  (
  
  courses2
  
  ,
  
  courses2_length
  
  ,
  
  sizeof
  
  (
  
  int
  
  ),
  
  compare
  
  );
 
 
 
 
  
  
  for 
  
  (
  
  i
  
  =
  
  0
  
  ,
  
  j
  
  =
  
  0
  
  ; 
  
  i
  
  <
  
  courses1_length 
  
  && 
  
  j
  
  <
  
  courses2_length
  
  ; 
  
  ) 
  
  {
 
 
 
 
      
  
  if 
  
  (
  
  courses1
  
  [
  
  i
  
  ]
  
  ==
  
  courses2
  
  [
  
  j
  
  ]) 
  
  {
 
 
 
 
          
  
  int 
  
  num  
  
  = 
  
  courses1
  
  [
  
  i
  
  ];
 
 
 
 
          
  
  printf
  
  (
  
  "%d
  
  \n
  
  "
  
  ,
  
  courses1
  
  [
  
  i
  
  ]);
 
 
 
 
          
  
  while 
  
  (
  
  courses1
  
  [
  
  ++
  
  i
  
  ]
  
  ==
  
  num
  
  );
 
 
 
 
          
  
  while 
  
  (
  
  courses2
  
  [
  
  ++
  
  j
  
  ]
  
  ==
  
  num
  
  );
 
 
 
 
      
  
  } 
  
  else 
  
  if 
  
  (
  
  courses1
  
  [
  
  i
  
  ]
  
  <
  
  courses2
  
  [
  
  j
  
  ]) 
  
  {
 
 
 
 
          
  
  i
  
  ++
  
  ;
 
 
 
 
      
  
  } 
  
  else 
  
  {
 
 
 
 
          
  
  j
  
  ++
  
  ;
 
 
 
 
      
  
  }
 
 
 
 
  
  
  }
 
 
 
 

  
  }
 
 
 
 

  
  （程序来源@www.talentbuddy.co）
 
 
 
 

  
  之所以贴这个程序是因为我在做这一题的时候显示寻找两个数组的公共元素，然后把公共元素提取出来放在额外建好的数组里
 
 
 
 

  
  在进行排序处理，最后输出。但是显然我的想法比这个算法的时间和空间复杂度都要高，这是我应该学习的