为类创建构造函数时，不会为该类创建任何默认构造函数。因此，如果你扩展该类，并且如果子类试图调用其超类的no-arg构造函数，则会出现编译时错误。

when you create a constructor for a class, there won't be any default constructor created for that class. so if you extend that class and if the subclass tries to call the no-arg constructor of its super class then there will be an compile-time error.

来演示：

class Parent {
  int i;
  public Parent(int i) {
    this.i=i;
  }
}

class Child extends Parent {
  int j;
  public Child(int i, int j) {
    super(i);
    this.j=j;
  }
  public Child(int j) {
    // here a call to super() is made, but since there is no no-arg constructor
    // for class Parent there will be a compile time error
    this.j=j;
  }
}

编辑：

回答你的问题，不要将值 arrLength 分配给 arr [] 和检查[] ，因为当时arrLength将是 0 。

to answer your question do this, don't assign the value arrLength to arr[] and check[] as arrLength would be 0 at that time.

所以只需要声明它们

int arr[];
boolean check[];

并在将输入分配给 arrLength $ c后的构造函数中$ c>放这些陈述。

and in the constructor after you assign the input to arrLength put these statements.

arr = new int[arrLength];
check = new boolean [arrLength];