Java实现算法导论中快速傅里叶变换FFT递归算法

要结合算法导论理解，参考：http://blog.youkuaiyun.com/fjssharpsword/article/details/53281889

代码中算法思路：输入n位（2的幂）向量，分别求值FFT和插值逆FFT，并计算卷积。

package sk.mlib;
/*************************************************************************
 *  Compilation:  javac FFT.java
 *  Execution:    java FFT N
 *  Dependencies: Complex.java
 *
 *  Compute the FFT and inverse FFT of a length N complex sequence.
 *  Bare bones implementation that runs in O(N log N) time. Our goal
 *  is to optimize the clarity of the code, rather than performance.
 *
 *  Limitations
 *  -----------
 *   -  assumes N is a power of 2
 *
 *   -  not the most memory efficient algorithm (because it uses
 *      an object type for representing complex numbers and because
 *      it re-allocates memory for the subarray, instead of doing
 *      in-place or reusing a single temporary array)
 *  
 *************************************************************************/
public class FFT {
	// compute the FFT of x[], assuming its length is a power of 2
    public static Complex[] fft(Complex[] x) {
        int N = x.length;

        // base case
        if (N == 1) return new Complex[] { x[0] };

        // radix 2 Cooley-Tukey FFT
        if (N % 2 != 0) { throw new RuntimeException("N is not a power of 2"); }

        // fft of even terms
        Complex[] even = new Complex[N/2];
        for (int k = 0; k < N/2; k++) {
            even[k] = x[2*k];
        }
        Complex[] q = fft(even);

        // fft of odd terms
        Complex[] odd  = even;  // reuse the array
        for (int k = 0; k < N/2; k++) {
            odd[k] = x[2*k + 1];
        }
        Complex[] r = fft(odd);

        // combine
        Complex[] y = new Complex[N];
        for (int k = 0; k < N/2; k++) {
            double kth = -2 * k * Math.PI / N;
            Complex wk = new Complex(Math.cos(kth), Math.sin(kth));
            y[k]       = q[k].plus(wk.times(r[k]));
            y[k + N/2] = q[k].minus(wk.times(r[k]));
        }
        return y;
    }


    // compute the inverse FFT of x[], assuming its length is a power of 2
    public static Complex[] ifft(Complex[] x) {
        int N = x.length;
        Complex[] y = new Complex[N];

        // take conjugate
        for (int i = 0; i < N; i++) {
            y[i] = x[i].conjugate();
        }

        // compute forward FFT
        y = fft(y);

        // take conjugate again
        for (int i = 0; i < N; i++) {
            y[i] = y[i].conjugate();
        }

        // divide by N
        for (int i = 0; i < N; i++) {
            y[i] = y[i].scale(1.0 / N);
        }

        return y;

    }

    // compute the circular