斐波那契数列

long long Fibonacci(unsigned int n){    if(n<=0)        return 0;    if(n==1)        return 1;    return Fibonacci(n-1)+Fibonacci(n-2);}//面试官期待的算法,时间复杂度降了long long Fibonacci(unsiged n){    int result[2] = {0,1};    if(n<2)    {        return result[n];    }    long long fibNMinusOne = 1;    long long fibNMinusTwo = 0;    for (unsigned int i=2;i<n;++i)    {        fibN = fibNMinusOne+fibNMinusTwo;        fibNMinusTwo = fibNMinusOne;        fibNMinusOne = fibN;    }    return fibN;}