只是为了使用异或来回答一个相对简单的问题……

return a ^ b ? c : a

我找到了这个解。

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    bool result = !(a ^ b ^ c) && !(!a & !b & !c) || (a & b & c);
    return result;
}

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?：

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.

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。

function atLeastTwoTrue($a, $b, $c) {

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