Question

我的代码有问题。我的目标是将伪代码转换为java代码是的，我的编码是一项任务，我不想要任何答案只是为了告诉我问题在哪里

我要做的是计算两个未分类的学生列表中没有重复项的交集的大小。

我将显示与此伪代码对应的伪代码和我的java代码。

伪代码：

inter <-- 0

Array C[m+n]

for i <-- 0 to m-1  do  C[i] <-- A[i]

for i <-- 0 to n-1  do  C[i+m] <-- B[i]

C <-- sort(C, m+n);

pointer <-- 0

while (pointer < m+n-1) do{

if(C[pointer]=C[pointer+1]){

inter <-- inter+1

pointer <-- pointer+2

}

else pointer <-- pointer+1

}

return inter

Java代码：

public static int intersectionSizeMergeAndSort(studentList L1, studentList L2) {
/* Write your code for question 4 here */
  int intersectionSize = 0;
  int[] C = new int[L1.studentID.length+L2.studentID.length];
  for(int i = 0; i<L1.studentID.length; i++){
  C[i] = L1.studentID[i];
  }
  for(int i = 0; i<L2.studentID.length; i++){
  C[i+L1.studentID.length] = L2.studentID[i];
  }
  Arrays.sort(C);
  int pointer = 0;
  while(pointer<((C.length-1))){
    if(C[pointer] == C[pointer+1]){
    intersectionSize = intersectionSize + 1;
    pointer = pointer + 2;
    }
    else {
      pointer = pointer + 1;
  }
 return intersectionSize;

}
return 0;
}

我的主要方法：

public static void main(String args[]) throws Exception {

studentList firstList;
studentList secondList;

// This is how to read lists from files. Useful for debugging.

// firstList=new studentList("COMP250.txt", "COMP250 - Introduction to Computer Science");
// secondList=new studentList("MATH240.txt", "MATH240 - Discrete Mathematics");

// get the time before starting the intersections
long startTime = System.currentTimeMillis();

// repeat the process a certain number of times, to make more accurate average     measurements.
 for (int rep=0;rep<1000;rep++) {

 // This is how to generate lists of random IDs. 
 // For firstList, we generate 16000 IDs
 // For secondList, we generate 16000 IDs

 firstList=new studentList(2 , "COMP250 - Introduction to Computer Science");
 secondList=new studentList(2 , "MATH240 - Discrete Mathematics");


 // run the intersection method
 int intersection=studentList.intersectionSizeMergeAndSort(firstList,secondList);
 System.out.println("The intersection size is: "+intersection);
 }

// get the time after the intersection
long endTime = System.currentTimeMillis();


System.out.println("Running time: "+ (endTime-startTime) + " milliseconds");
 }

}

注意：先前已声明L1和L2 但我没有得到我的目标。有人可以指出出了什么问题吗？

谢谢

Answer 1

在我的脑海中，在我看来你的return intersectionSize;语句出现在你的while循环中，所以你循环永远不会超出第一次迭代并且没有正确计算intersectionSize。我会删除该声明，并将return 0;替换为return intersectionSize;，如此...

public static int intersectionSizeMergeAndSort(studentList L1, studentList L2) {
   /* Write your code for question 4 here */
   int intersectionSize = 0;
   int[] C = new int[L1.studentID.length + L2.studentID.length];
   for (int i = 0; i < L1.studentID.length; i++) {
      C[i] = L1.studentID[i];
   }
   for (int i = 0; i < L2.studentID.length; i++) {
      C[i + L1.studentID.length] = L2.studentID[i];
   }
   Arrays.sort(C);
   int pointer = 0;
   while (pointer < (C.length - 1)) {
       if (C[pointer] == C[pointer + 1]) {
          intersectionSize = intersectionSize + 1;
          pointer = pointer + 2;
       } else {
          pointer = pointer + 1;
       }
    }
    return intersectionSize;
}

Answer 2

看起来您使用studentList.StudentID的长度而不是studentList的长度。

查看您的算法，您应该写L1.length而不是L1.StudentID.length，L2也是如此。

编辑：

我刚看到你的编辑。您需要在主for loop之外取出以下行：

// run the intersection method
   int intersection=studentList.intersectionSizeMergeAndSort(firstList,secondList);
   System.out.println("The intersection size is: "+intersection);

要正确记录您的时间，您需要在循环之后和我之前提到的前一行之前放置行long startTime = System.currentTimeMillis();。

使用Java的这种算法的伪代码

2 个答案: