面试题10：斐波那契数列

求斐波那契数列的第项。

答案：

// ====================方法1：递归====================
long long Fibonacci_Solution1(unsigned int n)
{
    if (n==0) return 0;
    else if (n==1) return 1;
    else return Fibonacci_Solution1(n-1)+Fibonacci_Solution1(n-2);
}

// ====================方法2：循环递推====================
long long Fibonacci_Solution2(unsigned int n)
{
    if (n==0) return 0;
    else if (n==1) return 1;
    else
    {
        long long prepre = 0, pre = 1;
        long long ret = 0;
        for (int i=2; i<=n; i++)
        {
            ret = pre + prepre;
            prepre = pre;
            pre = ret;
        }
        return ret;
    }
}

// =================方法3：基于矩阵乘法=================
#include <cassert>

struct Matrix2By2
{
    explicit Matrix2By2(long long m00 = 0, long long m01 = 0,
                        long long m10 = 0, long long m11 = 0)
            :m_00(m00), m_01(m01), m_10(m10), m_11(m11) {}

    long long m_00;
    long long m_01;
    long long m_10;
    long long m_11;
};

Matrix2By2 MatrixMultiply(const Matrix2By2& matrix1, const Matrix2By2& matrix2)
{
    return Matrix2By2(
            matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,
            matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,
            matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,
            matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);
}

Matrix2By2 MatrixPower(unsigned int n)
{
    assert(n > 0);

    Matrix2By2 matrix;
    if(n == 1)
    {
        matrix = Matrix2By2(1, 1, 1, 0);
    }
    else if(n % 2 == 0)
    {
        matrix = MatrixPower(n / 2);
        matrix = MatrixMultiply(matrix, matrix);
    }
    else if(n % 2 == 1)
    {
        matrix = MatrixPower((n - 1) / 2);
        matrix = MatrixMultiply(matrix, matrix);
        matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
    }

    return matrix;
}

long long Fibonacci_Solution3(unsigned int n)
{
    int result[2] = {0, 1};
    if(n < 2)
        return result[n];

    Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
    return PowerNMinus2.m_00;
}