题目九：斐波那契数列

Posted 2022-05-22 yzdai

/////////////////////////////////////////////////////////////////////////////////////////////
// 13.题目九：斐波那契数列
/*写一个函数，输入N,求斐波那契(Fibonacci)数列的第N项.
F(n) = 0    n == 0;
         1    n == 1;
         F(n -1) + F(n -2) n > 1;
        

*/

static int s_RunTime = 0;
void CalcFbio(int iNum)

    printf("N(%d) --> F(%d) + F(%d)\n", iNum, iNum - 1, iNum - 2);
    s_RunTime++;


int Fbio_1(int iNum)

    if (iNum <= 1)
    
        return iNum;
    

    CalcFbio(iNum);
    return Fbio_1(iNum - 1) + Fbio_1(iNum - 2);


// 优化版 --> 去掉重复计算项
//时间复杂度O(n)
int Fbio_2(int iNum, vector<int>& vect)

    // 1.循环退出条件
    if (iNum <= 1)
    
        return iNum;
    

    // 如果该项F(n)已经计算过，直接返回结果
    if (vect[iNum - 1] > 0)
    
        return vect[iNum - 1];
    

    // 3.存储对应项结果
    vect[iNum - 1] = Fbio_2(iNum - 1, vect) + Fbio_2(iNum - 2, vect);
    CalcFbio(iNum);
    return vect[iNum - 1];


// 非递归解法 时间复杂度O(n)
int Fbio_3(int iNum)

    int iRes = 0;
    if (iNum <= 1)
    
        return 1;
    

    int iSum = 0;
    int iOne = 1;
    int iTwo = 0;
    for (int i = 2; i <= iNum; i++)
    
        iSum = iOne + iTwo;
        iTwo = iOne;
        iOne = iSum;
        printf("Fbio: N = %d, Value = %d\n", i, iSum);
    

    return iSum;


//////////////////////////////////////////////////////////////////////////
// 类似题目：一只青蛙一次可以跳上一级台阶，也可以跳上2级，求该青蛙跳上一个N级的台阶总共有多少种跳法？
int JumpStep_1(int iNum)

    if (iNum <= 2)
    
        return iNum;
    

    CalcFbio(iNum);
    return JumpStep_1(iNum - 1) + JumpStep_1(iNum - 2);


int JumpStep_2(int iNum)

#if 0
    //int iSumStep = 0;
    int iStep1 = 1;
    int iStep2 = 2;

    while (iNum-- > 1)
    
        iStep2 += iStep1;
        iStep1 = iStep2 - iStep1;
        printf("JumpStep: N = %d, Value = %d\n", iNum, iStep1);
    

    return iStep1;

#else 

#endif

    if (iNum <= 2)
    
        return iNum;
    

    int iRes = 0;
    int iOne = 1;
    int iTwo = 2;

    for (int i = 3; i <= iNum; i++)
    
        iRes = iOne + iTwo;
        iOne = iTwo;
        iTwo = iRes;
        printf("JumpStep: N = %d, Value = %d\n", i, iRes);
    

    return iRes;
  


void FiboTestFunc()

    cout << "\n\n --------------- FiboTestFunc Start -------------->" << endl;

    //for (int i = 0; i < 20; i++)
    //
    //    printf("斐波那契数列: N = %d, Value = %d\n", i, Fbio(i));
    //
    
    int iNum = 10;
    printf("斐波那契数列: N = %d, Value = %d\n", iNum, Fbio_1(iNum));
    printf("方法一运行次数: %d\n", s_RunTime);

    cout << "========================>>>>>>>>>>>>>>>>>>>>>>>>>>>>" << endl;
    vector<int> vect(iNum, 0);
    s_RunTime = 0;
    printf("斐波那契数列: N = %d, Value = %d\n", iNum, Fbio_2(iNum, vect));
    printf("方法二运行次数: %d\n", s_RunTime);

    cout << "========================>>>>>>>>>>>>>>>>>>>>>>>>>>>>" << endl;
    printf("斐波那契数列: N = %d, Value = %d\n", iNum, Fbio_3(iNum));

    cout << "========================>>>>>>>>>>>>>>>>>>>>>>>>>>>>" << endl;
    s_RunTime = 0;
    printf("青蛙跳台阶: N = %d, Value = %d\n", iNum, JumpStep_1(iNum));
    printf("方法一运行次数: %d\n", s_RunTime);

    cout << "========================>>>>>>>>>>>>>>>>>>>>>>>>>>>>" << endl;
    printf("青蛙跳台阶: N = %d, Value = %d\n", iNum, JumpStep_2(iNum));

    cout << "\n\n --------------- FiboTestFunc End -------------->" << endl;