int Fibonacci(int n)
{
   if (1 == n)
   {
       return 1;
   }
   if (2 == n)
   {
       return 2;
   }
   return Fibonacci(n-1) + Fibonacci(n-2);
}
int FibonacciIteration (int n)
{</p>

<p>int a = 1;
   int b = 2;
   if (n == 1)
       return a;
   if (n == 2)
       return b;
   for (int i = 3;i    {
       int now = a + b;
       a = b;
       b = now;
   }
   return b;
}

书中提到了一种方法： fi = fi-1 + fi-2 =>转换成矩阵的形式

[编程之美_2.9]斐波那契数列
时间复杂度为：O(log2n)