一位面试官最近问了我这样一个问题:给定三个布尔变量a、b和c,如果三个变量中至少有两个为真,则返回true。

我的解决方案如下:

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    if ((a && b) || (b && c) || (a && c)) {
        return true;
    }
    else{
        return false;
    }
}

他说这还可以进一步改进,但如何改进呢?


当前回答

我相信使用普通的布尔运算符(a || b) && (b || c)是很好的,而且更简单。

你可以把这三个字母中的任意一个和另外两个字母中的任意一个交换它仍然是相同的表达式。

其他回答

而不是写:

if (someExpression) {
    return true;
} else {
    return false;
}

写:

return someExpression;

至于表达式本身,是这样的:

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    return a ? (b || c) : (b && c);
}

或者这个(你觉得哪个更容易理解):

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    return a && (b || c) || (b && c);
}

它只测试a和b一次,c最多测试一次。

参考文献

JLS 15.25条件运算符?:

当我看到这个问题时,我的第一个想法是:

int count=0;
if (a)
    ++count;
if (b)
    ++count;
if (c)
    ++count;
return count>=2;

在看了其他帖子后,我承认

return (a?1:0)+(b?1:0)+(c?1:0)>=2;

更优雅。我想知道相对运行时是什么。

不过,无论如何,我认为这种解决办法比那种解决办法要好得多

return a&b | b&c | a&c;

variety because is is more easily extensible. What if later we add a fourth variable that must be tested? What if the number of variables is determined at runtime, and we are passed an array of booleans of unknown size? A solution that depends on counting is much easier to extend than a solution that depends on listing every possible combination. Also, when listing all possible combinations, I suspect that it is much easier to make a mistake. Like try writing the code for "any 3 of 4" and make sure you neither miss any nor duplicate any. Now try it with "any 5 of 7".

One thing I haven't seen others point out is that a standard thing to do in the "please write me some code" section of the job interview is to say "Could you improve that?" or "Are you completely happy with that" or "is that as optimized as possible?" when you say you are done. It's possible you heard "how would you improve that" as "this might be improved; how?". In this case changing the if(x) return true; else return false; idiom to just return x is an improvement - but be aware that there are times they just want to see how you react to the question. I have heard that some interviewers will insist there is a flaw in perfect code just to see how you cope with it.

您不需要使用运算符的短路形式。

返回(a & b) | (b & c) | (c & a);

它执行与您的版本相同数量的逻辑操作,但是完全没有分支。

另一个直接代码的例子:

int  n = 0;
if (a) n++;
if (b) n++;
if (c) n++;
return (n >= 2);

显然,这不是最简洁的代码。

齿顶高

另一个(稍微优化的)版本:

int  n = -2;
if (a) n++;
if (b) n++;
if (c) n++;
return (n >= 0);

这可能会运行得稍微快一点,假设与0的比较将比与2的比较使用更快(或更少)的代码。