java - 顺序搜索 Java

Question

我在按顺序搜索时遇到问题，这是练习:

You are testing the physical endurance of a type of brick. The test is made to find the max height the brick can fall without breaking from and N stories building. You have k bricks and two cases:

Case 1. When k = 1 the tries begin from the current floor upwards, sequentially, until the brick breaks.

Case 2. For k > 1 one brick is dropped from floor N/2. If it breaks, case 2 is applied again between floors 1 and N/2 -1. If the brick does not break, case 2 is applied again between floors N/2 + 1 and N.

我对案例 1 的原因有疑问，因为我不知道如何处理它:

Initially you call dropBricks(k,1,N); Being k, and N introduced by the User(Keyboard).

public static boolean breakingFloor(int floor){
    boolean breaks;
    Random r =new Random();
    int breakingFloor= r.nextInt(floor)+1;
    
    if(floor>=breakingFloor){
        breaks=true;
    }else{
        breaks=false;
    }
    
  return breaks;
}

dropBricks receives k(number of remaining bricks, first (first floor),last (last floor = N)) and the output would be the breakingFloor

public static int dropBricks(int k,int first,int last){
    int result=0;
    
    if (k==1){//CASE 1
        //I KNOW THIS CODE IS WRONG, IS JUST O NOT LEAVE IT IN EMPTY
         do{
             dropBricks(k,first,last);
             first++;
         }while(breakingFloor(first));
         System.out.println("Floor brick broke "+floor);
   
    }else{//CASE 2
        if(last-first<=1){
            if(last==first){
                result=first;
                System.out.println("Floor brick broke "+result);
            }else{
                if(breakingFloor(first)){
                    result=first;
                    k--;
                    System.out.println("Floor brick broke "+result);
                }else{
                    result=last;
                    k--;
                    System.out.println("Floor brick broke "+result);
                }
            }
        }else{
            int mid=((first+last)/2);
            if(breakingFloor(mid)){
                result=dropBricks(k-1,first,mid);
                System.out.println("Floor brick broke "+result);
            }else{
                result=dropBricks(k,mid,last);
                System.out.println("Floor brick broke "+result);
            }
        }      
    }
    return result;
}

Answer 1

您的问题 - 或者至少其中之一 - 是每次您绕过这个循环...

do{
    dropBricks(k,first,last);
    first++;
} while( breakingFloor(first) );

...您正在选择一个新的随机楼层，它会在该楼层破裂(发生在 breakingFloor 方法中):

Random r =new Random();
int breakingFloor= r.nextInt(floor)+1;

这肯定会给您带来非常不一致的行为。它有时可能真的会侥幸成功。

您想选择它在开始时会破裂的地板并使其保持不变。

测试您的应用程序的明智方法可能是先对断层进行硬编码，然后再引入随机性。