《剑指offer》第十题（斐波那契数列）

// 面试题：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 <iostream>

using namespace std;

// ====================方法1：递归====================
//注意这种递归方法虽然看起来很简单，但是由于压入栈和弹出，会存在栈溢出的可能，而且效率特别慢，且n越大效率越慢
long long Fibonacci_Solution1(unsigned int n)//注意long long
{
    if (n <= 0)
        return 0;

    if (n == 1)
        return 1;

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

// ====================方法2：循环====================
//这是一种简单的方法，时间复杂度O（n），值得提倡
long long Fibonacci_Solution2(unsigned n)
{
    int result[2] = { 0, 1 };//注意头两个数我们无法计算，直接给出
    if (n < 2)
        return result[n];

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

        fibNMinusTwo = fibNMinusOne;
        fibNMinusOne = fibN;
    }

    return fibN;
}

// ====================方法3：基于矩阵乘法====================
//不常见但是时间复杂度更低： O（logn）

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;
    }
    //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) {}
    //也可以写成上面这个形式，都是含参具有默认值的构造函数，是参数列表外初始化

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

Matrix2By2 MatrixMultiply(const Matrix2By2& matrix1,const Matrix2By2& matrix2)//矩阵乘法
{
    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)
{
    Matrix2By2 matrix;
    if (n == 1)//第一种情况：n=1，返回矩阵(1, 1, 1, 0)
    {
        matrix = Matrix2By2(1, 1, 1, 0);
    }
    else if (n % 2 == 0)//第二种情况：n为偶数，递归求a^(n/2)，然后乘回来
    {
        matrix = MatrixPower(n / 2);
        matrix = MatrixMultiply(matrix, matrix);
    }
    else if (n % 2 == 1)//第三种情况：n为奇数数，用第二种情况递归求a^((n-1)/2)，然后乘回来后，再乘一次(1, 1, 1, 0)
    {
        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 };//注意头两个数我们无法计算，直接给出
    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);
    else
        printf("Test for %d in solution1 failed.
", n);

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

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

    system("pause");
}

 第三种想法思路：

斐波那契数列扩展：