剑指Offer系列-面试题9：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。

效率很低的解法(递归实现)：

long long Fibonacci(unsigned int n){    if(n <= 0)        return 0;    if(n == 1)        return 1;    return Fibonacci(n - 1) + Fibonacci(n - 2);}

从f(0)、f(1)、f(2)由低往高算：

#include <iostream>using namespace std;long long Fibonacci(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;}int main(){    cout << Fibonacci(3) << endl;    cout << Fibonacci(5) << endl;    cout << Fibonacci(10) << endl;    return 0;}