Question

我很难尝试使用逻辑规则并使用OutOfOrder代码来使InOrder像OurOfOrder一样简单。

public class InOrder {
//Don't change this
public boolean OutOfOrder(int n1, int n2, int n3) {
return (n1 > n2) || (n2 > n3);
}

//The original and messy InOrder, leave this as an example of what not to do
public boolean inOrder(int n1, int n2, int n3) {
if (n2 > n1) {
  if (n3 > n2) {
    return true;
  } else {
    return false;
  }
} else if (n2 == n1) {
  if (n3 == n1) {
    return true;
  } else {
    return false;
  }
} else {
  return false;
}
}

//The new and improved InOrder for part 5, call OutOfOrder
public boolean inOrder5a(int n1, int n2, int n3) {
return true; //replace this
}

//The newer and improved InOrder for part 6, inline
public boolean inOrder5b(int n1, int n2, int n3) {
return true;//replace this
}

}

Answer 1

数字似乎是有序或无序的，没有第三种选择。

因此您可以将inOrder实施为：

public boolean inOrder(int n1, int n2, int n3) {
    return !outOfOrder(n1, n2, n3);
}

Answer 2

我会说：(n2 >= n1 && n3 >= n2)

有没有一种方法可以使用OutOfOrder的代码，使InOrder像OutOfOrder一样简单？

2 个答案: