java - 当存在 2^n 条件时从递归函数返回的更好方法

Question

假设我有n = 3 boolean变量 -- a , b和c在Java .

这些变量总共可以有 8 (2^n = 2^3)组合。

这些条件用于确定return recursive 中的声明功能如下:

static int recursiveFunc(int x){

    boolean a, b, c;

    a = getBoolVal(x);
    b = getBoolVal(x + 1);
    c = getBoolVal(x + 2);

    if(a == true && b == true && c == true)        //7
        return recursiveFunc(x + 1) + recursiveFunc(x + 2) + recursiveFunc(x + 3);

    else if(a == true && b == true && c == false)  //6
        return recursiveFunc(x + 1) + recursiveFunc(x + 2);

    else if(a == true && b == false && c == true)  //5
        return recursiveFunc(x + 1) + recursiveFunc(x + 3);

    else if(a == true && b == false && c == false) //4
        return recursiveFunc(x + 1);

    else if(a == false && b == true && c == true)  //3
        return recursiveFunc(x + 2) + recursiveFunc(x + 3);

    else if(a == false && b == true && c == false) //2
        return recursiveFunc(x + 2);

    else if(a == false && b == false && c == true) //1
        return recursiveFunc(x + 3);

    else                                           //0
        return 0;

}

static boolean getBoolVal(int x){

    if(some condition with respect to x)
        return true;

    else
        return false;

}

正如您所看到的，随着n值的增加，条件的数量将变得相当长。 .

但是，可以在 n + 2 中轻松生成返回语句。步骤(而不是 2^n )如下:

String returnStat = "";

if(a == true)
    returnStat += "recursiveFunc(x + 1) + ";

if(b == true)
    returnStat += "recursiveFunc(x + 2) + ";

if(c == true)
    returnStat += "recursiveFunc(x + 3) + ";

if(returnStat == "")
    returnStat = "0";

else
    returnStat = returnStat.substring(0, returnStat.length() - 3); //removing extra " + "

无论如何我可以返回returnStat作为一个声明？也许，类似——

return stringToCode(returnStat);

如果没有，如何克服这种情况？

Answer 1

recursiveFunc 返回一个 int，因此只需将 int 加起来即可:

int result = 0;

if (a)
    result += recursiveFunc(x + 1);
if (b)
    result += recursiveFunc(x + 2);
if (c)
    result += recursiveFunc(x + 3);

return result;