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".

在Clojure中:

(defn at-least [n & bools]
  (>= (count (filter true? bools)) n)

用法:

(at-least 2 true false true)

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?：

我不喜欢三元(return a ?(b || c):(b && c);从最上面的答案)，我想我没有看到任何人提到过它。它是这样写的:

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

public static boolean atLeast(int atLeastToBeTrue, boolean...bools){
    int booleansTrue = 0;
    for(boolean tmp : bools){
        booleansTrue += tmp ? 1 : 0;
    }
    return booleansTrue >= atLeastToBeTrue;
}

你可以从varargs a.k.a boolean[]中选择至少有多少你想为真:-)

问题中的2和3显然是神奇的数字。“正确”的答案将取决于面试官是否试图了解你对布尔逻辑的理解(我不认为pdox的答案在这方面是最好的)或你对架构问题的理解。

我倾向于使用映射减少解决方案，它将接受任何类型的列表和任何任意条件。