剑指offer——面试题10：斐波那契数列

个人答案：

#include"iostream"
#include"stdio.h"
#include"string.h"
using namespace std;
typedef long long ll;
const int MAXN=10000;

ll fib[MAXN];
ll Fibonacci(int n)
{
    if(fib[n]!=-1)
        return fib[n];
    return fib[n]=Fibonacci(n-1)+Fibonacci(n-2);
}

int main()
{
    int n;
    memset(fib,-1,sizeof(fib));
    fib[0]=0;
    fib[1]=1;
    while(cin>>n)
    {
        cout<<Fibonacci(n)<<endl;
    }
    return 0;
}

官方答案：

// 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====================
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====================
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(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====================
#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)
    {
        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;
}

// ====================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);

    return 0;
}