斐波那契数列递归解法与循环解法

写一个函数，输入n，求斐波那契数列的第n项。

递归解法(效率很低):

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

循环解法(O(n)):

long long Fibonacci( unsigned n )
{
	int result[2] = { 0, 1 };

	if ( n < 2 )
		return result[n];

	long long fibNMinusOne = 0;
	long long fibNMinusTwo = 1;
	long long fibN = 0;

	for ( unsigned i = 2; i <= n; ++i ){
		fibN = fibNMinusOne + fibNMinusTwo;

		fibNMinusOne = fibNMinusTwo;
		fibNMinusTwo = fibN;
	}

	return fibN;
}