使用Strassen算法求两个2^k×2^k的矩阵的乘积的java代码，并分析算法的计算时间复杂性.

以下是使用Strassen算法求两个2^k×2^k的矩阵的乘积的Java代码：

public class StrassenMatrixMultiplication {
    public static int[][] multiply(int[][] A, int[][] B) {
        int n = A.length;
        int[][] C = new int[n][n];
        if (n == 1) {
            C[0][0] = A[0][0] * B[0][0];
        } else {
            int[][] A11 = new int[n/2][n/2];
            int[][] A12 = new int[n/2][n/2];
            int[][] A21 = new int[n/2][n/2];
            int[][] A22 = new int[n/2][n/2];
            int[][] B11 = new int[n/2][n/2];
            int[][] B12 = new int[n/2][n/2];
            int[][] B21 = new int[n/2][n/2];
            int[][] B22 = new int[n/2][n/2];

            // Divide matrices A and B into 4 sub-matrices
            divide(A, A11, 0 , 0);
            divide(A, A12, 0 , n/2);
            divide(A, A21, n/2, 0);
            divide(A, A22, n/2, n/2);
            divide(B, B11, 0 , 0);
            divide(B, B12, 0 , n/2);
            divide(B, B21, n/2, 0);
            divide(B, B22, n/2, n/2);

            // Calculate 7 products recursively
            int[][] P1 = multiply(add(A11, A22), add(B11, B22));
            int[][] P2 = multiply(add(A21, A22), B11);
            int[][] P3 = multiply(A11, sub(B12, B22));
            int[][] P4 = multiply(A22, sub(B21, B11));
            int[][] P5 = multiply(add(A11, A12), B22);
            int[][] P6 = multiply(sub(A21, A11), add(B11, B12));
            int[][] P7 = multiply(sub(A12, A22), add(B21, B22));

            // Calculate sub-matrices of C
            int[][] C11 = add(sub(add(P1, P4), P5), P7);
            int[][] C12 = add(P3, P5);
            int[][] C21 = add(P2, P4);
            int[][] C22 = add(sub(add(P1, P3), P2), P6);

            // Combine sub-matrices of C into C
            copy(C11, C, 0 , 0);
            copy(C12, C, 0 , n/2);
            copy(C21, C, n/2, 0);
            copy(C22, C, n/2, n/2);
        }
        return C;
    }

    // Divide matrix A into 4 sub-matrices
    public static void divide(int[][] A, int[][] B, int iB, int jB) {
        for(int i1 = 0, i2=iB; i1<B.length; i1++, i2++)
            for(int j1 = 0, j2=jB; j1<B.length; j1++, j2++)
                B[i1][j1] = A[i2][j2];
    }

    // Add matrices A and B
    public static int[][] add(int[][] A, int[][] B) {
        int n = A.length;
        int[][] C = new int[n][n];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                C[i][j] = A[i][j] + B[i][j];
        return C;
    }

    // Subtract matrices A and B
    public static int[][] sub(int[][] A, int[][] B) {
        int n = A.length;
        int[][] C = new int[n][n];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                C[i][j] = A[i][j] - B[i][j];
        return C;
    }

    // Copy matrix A into matrix B
    public static void copy(int[][] A, int[][] B, int iB, int jB) {
        for(int i1 = 0, i2=iB; i1<A.length; i1++, i2++)
            for(int j1 = 0, j2=jB; j1<A.length; j1++, j2++)
                B[i2][j2] = A[i1][j1];
    }

    public static void main(String[] args) {
        int[][] A = { {1, 2}, {3, 4} };
        int[][] B = { {5, 6}, {7, 8} };
        int[][] C = multiply(A, B);
        for (int i = 0; i < C.length; i++) {
            for (int j = 0; j < C.length; j++) {
                System.out.print(C[i][j] + " ");
            }
            System.out.println();
        }
    }
}

Strassen算法的时间复杂度为O(n^log2 7)，其中n是矩阵的大小。在这里，n是2^k，因此时间复杂度为O(2^(k*log2 7))，即O(n^2.81)。在比较常见的情况下，Strassen算法的运行时间比传统的矩阵乘法算法要快，但对于较小的矩阵，传统的算法可能更快。