Question

我有顺序搜索的问题，这是练习：

您正在测试一种砖的物理耐力。测试是为了找到砖块可以在不破坏和N层建筑的情况下落下的最大高度。你有k砖和两个案例：

案例1.当k = 1时，尝试从当前楼层向上依次开始，直到砖块断裂。

案例2.对于k＆gt;从地板N / 2上掉下一块砖。如果它破裂，则在楼层1和N / 2 -1之间再次应用案例2。如果砖没有破裂，则在楼层N / 2 + 1和N之间再次应用案例2。

我对CASE 1的原因有疑问，因为我不知道如何处理它：

最初你调用dropBricks（k，1，N）;由用户（键盘）引入k和N.

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收到k（剩余砖块的数量，第一层（第一层），最后一层（最后一层= N）），输出将是破碎地板

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

你的问题 - 或者至少其中一个 - 就是每次你绕过这个循环......

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

...您正在选择一个新的随机区域，它将在其中断（在breakingFloor方法中发生）：

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

这肯定会给你非常不一致的行为。它可能实际上有些时候是侥幸。

您想要选择在开始时会破坏的楼层，并保持不变。

测试应用程序的一种明智方法可能是硬编码破土动工，然后再引入随机性。

Answer 2

您的第一个问题在这里：

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这根本没有意义。砖块断裂的楼层编号不随机。

换句话说：您想为该砖分配一个一次的随机数。然后你的算法必须找到那个数字！您的代码会更改＆＃34;破坏楼层编号＆＃34; 而您的算法会尝试查找该号码。这使得整个过程毫无用处（当然，＆＃34;随机＆＃34 ;;因为各种各样的事情都可能发生）。

所以迈出第一步：退一步，重新思考问题，解决问题＆＃34;问题真的要求你这样做。

顺序搜索Java

2 个答案: