《剑指offer》第十题：斐波那契数列

// 面试题10：斐波那契数列
// 题目：写一个函数，输入n，求斐波那契（Fibonacci）数列的第n项。

#include <cstdio>

// ====================方法1：递归====================
//存在函数调用导致栈溢出的可能, 效率低, 时间复杂度指数递增
long long Fibonacci_Solution1(unsigned int n)
{
    if (n <= 0)
        return 0;
    if (n == 1)
        return 1;

    return Fibonacci_Solution1(n - 1) + Fibonacci_Solution1(n - 2);
}

// ====================方法2：循环====================
//时间复杂度O(n)
long long Fibonacci_Solution2(unsigned n)
{
    int result[2] = { 0, 1 };
    if (n < 2)
        return result[n];

    long long fibNMinusOne = 1;
    long long fibNMinusTwo = 0;
    long long fibN = 0;
    for (int i = 2; i <= n; ++i)
    {
        fibN = fibNMinusOne + fibNMinusTwo;

        fibNMinusTwo = fibNMinusOne;
        fibNMinusOne = fibN;
    }
    return fibN;
}

// ====================方法3：基于矩阵乘法====================
//时间复杂度O(logn)
#include <cassert>

struct Matrix2By2
{
    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) //n为偶数, 递归求 a^(n/2), 然后平方
    {
        matrix = MatrixPower(n / 2);
        matrix = MatrixMultiply(matrix, matrix);
    }
    else if (n % 2 == 1) //n为奇数, 递归求 a^((n-1)/2), 平方后乘a
    {
        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;
}

// ====================测试代码====================
void Test(int n, int expected)
{
    if (Fibonacci_Solution1(n) == expected)
        printf("Test for %d in solution1 passed.\n", n);
    else
        printf("Test for %d in solution1 failed.\n", n);

    if (Fibonacci_Solution2(n) == expected)
        printf("Test for %d in solution2 passed.\n", n);
    else
        printf("Test for %d in solution2 failed.\n", n);

    if (Fibonacci_Solution3(n) == expected)
        printf("Test for %d in solution3 passed.\n", n);
    else
        printf("Test for %d in solution3 failed.\n", n);
}

int main(int argc, char* argv[])
{
    Test(0, 0);
    Test(1, 1);
    Test(2, 1);
    Test(3, 2);
    Test(4, 3);
    Test(5, 5);
    Test(6, 8);
    Test(7, 13);
    Test(8, 21);
    Test(9, 34);
    Test(10, 55);

    Test(40, 102334155);

    return 0;
}

分析：分解问题为小问题，自下而上解决。

标签：matrix,offer,long,斐波,Matrix2By2,Test,那契,matrix2,matrix1
来源： https://www.cnblogs.com/ZSY-blog/p/12547937.html