作为@TofuBeer TofuBeer精彩帖子的补充，考虑@pdox pdox的回答:

static boolean five(final boolean a, final boolean b, final boolean c)
{
    return a == b ? a : c;
}

再考虑一下它的分解版本，如"javap -c"所给出的:

static boolean five(boolean, boolean, boolean);
  Code:
    0:    iload_0
    1:    iload_1
    2:    if_icmpne    9
    5:    iload_0
    6:    goto    10
    9:    iload_2
   10:    ireturn

Pdox的答案编译成的字节代码比之前的任何答案都要少。它的执行时间与其他的相比如何?

one                5242 ms
two                6318 ms
three (moonshadow) 3806 ms
four               7192 ms
five  (pdox)       3650 ms

至少在我的电脑上，pdox的回答比@moonshadow moonshadow的回答稍微快一点，使得pdox的回答是最快的(在我的惠普/英特尔笔记本电脑上)。

If the goal is to return a bitwise two-out-of-three value for three operands, arithmetic and iterative approaches are apt to be relatively ineffective. On many CPU architectures, a good form would be "return ((a | b) & c) | (a & b);". That takes four boolean operations. On single-accumulator machines (common in small embedded systems) that's apt to take a total of seven instructions per byte. The form "return (a & b) | (a & c) | (b & c);" is perhaps nicer looking, but it would require five boolean operations, or nine instructions per byte on a single-accumulator machine.

顺便提一下，在CMOS逻辑中，计算“不是三选二”需要12个晶体管(相比之下，逆变器需要2个晶体管，双输入NAND或NOR需要4个晶体管，而三输入NAND或NOR需要6个晶体管)。

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;
}

我想我还没见过这个解决方案:

boolean atLeast(int howMany, boolean[] boolValues) {
  // check params for valid values

  int counter = 0;
  for (boolean b : boolValues) {
    if (b) {
      counter++;

      if (counter == howMany) {
        return true;
      }
    }
  }
  return false;
}

它的优点是一旦达到你要找的数字，它就会崩溃。因此，如果这是“这1,000,000个值中至少有2个是真的”，而前两个值实际上是真的，那么它应该比一些更“正常”的解更快。

最明显的改进是:

// There is no point in an else if you already returned.
boolean atLeastTwo(boolean a, boolean b, boolean c) {
    if ((a && b) || (b && c) || (a && c)) {
        return true;
    }
    return false;
}

然后

// There is no point in an if(true) return true otherwise return false.
boolean atLeastTwo(boolean a, boolean b, boolean c) {
    return ((a && b) || (b && c) || (a && c));
}

但这些改进都是微不足道的。

设三个布尔值为A,B和C....

你可以使用一个k-MAP和一个布尔表达式…

在这种情况下，布尔表达式将是A(B+C) +C

或者if(A && (B || C) || C) ｛ 返回true; ｝ 其他的 返回错误;