函数ko返回答案:

static int ho(bool a)
{
    return a ? 1 : 0;
}

static bool ko(bool a, bool b, bool c)
{
    return ho(a) + ho(b) + ho(c) >= 2 ? true : false;
}

C解。

int two(int a, int b, int c) {
  return !a + !b + !c < 2;
}

或者你可能更喜欢:

int two(int a, int b, int c) {
  return !!a + !!b + !!c >= 2;
}

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".

return 1 << $a << $b << $c >= 1 << 2;

下面是使用map/reduce的另一个实现。在分布式环境中，这可以很好地扩展到数十亿布尔值©。使用MongoDB:

创建数据库的布尔值:

db.values.insert({value: true});
db.values.insert({value: false});
db.values.insert({value: true});

创建map, reduce函数:

编辑:我喜欢CurtainDog的回答有映射/减少适用于泛型列表，所以这里有一个地图函数，它接受一个回调，决定一个值是否应该被计数。

var mapper = function(shouldInclude) {
    return function() {
        emit(null, shouldInclude(this) ? 1 : 0);
    };
}

var reducer = function(key, values) {
    var sum = 0;
    for(var i = 0; i < values.length; i++) {
        sum += values[i];
    }
    return sum;
}

运行map / reduce:

var result = db.values.mapReduce(mapper(isTrue), reducer).result;

containsMinimum(2, result); // true
containsMinimum(1, result); // false


function isTrue(object) {
    return object.value == true;
}

function containsMinimum(count, resultDoc) {
    var record = db[resultDoc].find().next();
    return record.value >= count;
}