Practice it CSE142 MT 06au 的一道题

Write a method named hasMidpoint that accepts three integers as parameters and returns true if one of the integers is the midpoint between the other two integers; that is, if one integer is exactly halfway between them. Your method should return false if no such midpoint relationship exists.
The integers could be passed in any order; the midpoint could be the 1st, 2nd, or 3rd. You must check all cases.
Calls such as the following should return true:
hasMidpoint(4, 6, 8)
hasMidpoint(2, 10, 6)
hasMidpoint(8, 8, 8)
hasMidpoint(25, 10, -5)
Calls such as the following should return false:
hasMidpoint(3, 1, 3)
hasMidpoint(1, 3, 1)
hasMidpoint(21, 9, 58)
hasMidpoint(2, 8, 16)
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以下是我和晓天想出来的答案。我们最开始打算先进行排序,再求差,来判断,很是程序写起来会很长,而且我们现在没有讲各种排序法。
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# name expected return your return result
1 hasMidpoint(1, 2, 3) true true pass
2 hasMidpoint(2, 10, 6) true true pass
3 hasMidpoint(0, -50, -25) true true pass
4 hasMidpoint(21, 9, 58) false false pass
5 hasMidpoint(-2, 9, 27) false false pass
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提交答案之后,我突然觉得这个程序存在问题,虽然在系统给的数值里没有测出问题,但是由于int类数据的特点,10/3是等于3的。所以输入以下测试代码。
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//由于是复制过来的,就不能保持很好的代码格式了。

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输出结果:true
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很显然,-25不是-51和0的中点,刚刚的源码是有问题的。
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把if的条件语句作了如下调整:

再次调用

输出是false
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现在基本已经不会存在问题了,通过了我的所有测试。
如果有人认为我的代码还存在问题,欢迎回复。