java - 递归FFT Java算法返回null吗？

我目前正在尝试在Java中实现FFT算法，但遇到了一些麻烦!我已经对该算法的所有其他部分进行了很好的测试，并且它们似乎运行良好。

我遇到的麻烦是，在基本情况下，它返回了一个复数数组，在基本情况下填充了A[0]。执行完基本案例后，将执行for循环，其中y0[0]和y1[0]被发现为 null ，尽管将它们分配给了基本案例，对此感到很困惑。这在System.out.println行中显示

谁能告诉我我的方式的错误？

    //This method implements the recursive FFT algorithm, it assumes the input length
//N is some power of two
private static Complex[] FFT(Complex[] A, int N) {
    double real, imag;
    Complex A0[] = new Complex[((int) Math.ceil(N/2))];
    Complex A1[] = new Complex[((int) Math.ceil(N/2))];
    Complex[] omega = new Complex[N];
    Complex[] y = new Complex[N];
    Complex[] y0 = new Complex[((int) Math.ceil(N/2))];
    Complex[] y1 = new Complex[((int) Math.ceil(N/2))];

    //base case
    if (N == 1) {
        return A;
    }
    else {
        real = Math.cos((2*Math.PI)/N); if (real < 1E-10 && real > 0) real = 0;
        imag = Math.sin((2*Math.PI)/N); if (imag < 1E-10 && imag > 0) imag = 0;
        omega[N-1] = new Complex(real, imag);
        omega[0] = new Complex(1, 0);
        A0 = splitInput(A, 1);
        A1 = splitInput(A, 0);
        //recursive calls
        y0 = FFT(A0, N/2);
        y1 = FFT(A1, N/2);
        for (int k = 0; k < ((N/2)-1); k++) {
            System.out.print("k: " + k + ", y0: " + y0[k]); System.out.println(", y1: " + y1[k]);
            y[k] = y0[k].plus(omega[k].times(y1[k]));
            y[k+(N/2)] = y0[k].minus(omega[k].times(y1[k]));
            omega[0] = omega[0].times(omega[N]);
        }
        return y;
    }
}

这是请求的我的splitInput方法的代码

//This method takes a double array as an argument and returns every even or odd
//element according to the second int argument being 1 or 0
private static Complex[] splitInput(Complex[] input, int even) {
    Complex[] newArray = new Complex[(input.length/2)];

    //Return all even elements of double array, including 0
    if (even == 1) {
        for (int i = 0; i < (input.length/2); i++) {
            newArray[i] = new Complex(input[i*2].re, 0.0);
        }
        return newArray;
    }
    //Return all odd elements of double array
    else {
        for (int i = 0; i < (input.length/2); i++) {
            newArray[i] = new Complex (input[(i*2) + 1].re, 0.0);
        }
    return newArray;
    }
}

编辑:我已经按照您的建议更新了代码，仍然从y[k] = y0[k].plus(omega[k].times(y1[k]));行中得到了空指针异常，因为在基本情况之后y0和y1仍然是null :(还有其他想法吗？这是更新的算法

//This method implements the recursive FFT algorithm, it assumes the input length
//N is some power of two
private static Complex[] FFT(Complex[] A, int N) {
    double real, imag;
    Complex[] omega = new Complex[N];
    Complex[] y = new Complex[N];
    Complex[] A0;
    Complex[] A1;
    Complex[] y0;
    Complex[] y1;

    //base case
    if (N == 1) {
        return A;
    }
    else {
        real = Math.cos((2*Math.PI)/N); if (real < 1E-10 && real > 0) real = 0;
        imag = Math.sin((2*Math.PI)/N); if (imag < 1E-10 && imag > 0) imag = 0;;
        omega[N-1] = new Complex(real, imag);
        omega[0] = new Complex(1, 0);
        A0 = splitInput(A, 1);
        A1 = splitInput(A, 0);
        //recursive calls
        y0 = FFT(A0, N/2);
        y1 = FFT(A1, N/2);
        for (int k = 0; k < ((N/2)-1); k++) {
            y[k] = y0[k].plus(omega[k].times(y1[k]));
            y[k+(N/2)] = y0[k].minus(omega[k].times(y1[k]));
            omega[0] = omega[0].times(omega[N-1]);
        }
        return y;
    }
}

最佳答案

跳出来的问题:

(int) Math.ceil(N/2)您仍在进行int划分，因此Math.ceil()没有任何效果，并且您的拆分数组可能不适用于奇数n

您只曾经填充omega[0]和omega[N-1]，当您尝试访问NullPointerException时，我希望有一个omega[1]，这会在N >= 6时发生。

omega[N]，如sarnold

所述

您分配A0和A1，然后将它们分配给splitInput