C# 复杂递归函数

主程序

public class look
{
    public int takeALook (int a)
    {
        if (a == 1)
            return 1;
        else if (a == 0)
            return 0;
        else
            return takeALook(a-2) + takeALook(a-1);

    }
}

---

答案是8。但谁能解释一下原因吗？它把我弄糊涂了，因为它称自己为2x。

它归结起来如下：

int a = 6;

look lk = new look();

int r = lk.takeALook(a);

Console.WriteLine("r is" + r);

---

下面是一个以树的形式显示递归的小图片。修剪已经计算过的孩子们

只是用一个跟踪程序修改了你的程序，这个跟踪程序对于任何递归函数来说都非常方便

takeALook(0) == 0
takeALook(1) == 1
takeALook(2) == takeALook(0) + takeALook(1) == 0 + 1 == 1
takeALook(3) == takeALook(1) + takeALook(2) == 1 + 1 == 2
takeALook(4) == takeALook(2) + takeALook(3) == 1 + 2 == 3
takeALook(5) == takeALook(3) + takeALook(4) == 2 + 3 == 5
takeALook(6) == takeALook(4) + takeALook(5) == 3 + 5 == 8

结果:

const string space = "  ";
static string StrMultiplier(string str,int multiplier)
{
    return string.Concat(Enumerable.Repeat(str,multiplier).ToArray());
}

static int F(int a,int level = 0)
{
    Console.WriteLine("{0}->F({1})",StrMultiplier(space,level),a);//Trace line
    if (a == 1)
        return 1;
    else if (a == 0)
        return 0;
    else
        return F(a-2,level + 1) + F(a-1,level+1);

}

---

带上debugger看看你自己。你没听说过斐波那契数吗：你所做的就是写斐波那契序列。@Konstantin是的，我知道斐波那契序列，但我对函数本身的逻辑以及如何使其易于理解感到困惑。我试着调试，但不明白。太好了！谢谢你真的知道如何让它简单易懂：）

const string space = "  ";
static string StrMultiplier(string str,int multiplier)
{
    return string.Concat(Enumerable.Repeat(str,multiplier).ToArray());
}

static int F(int a,int level = 0)
{
    Console.WriteLine("{0}->F({1})",StrMultiplier(space,level),a);//Trace line
    if (a == 1)
        return 1;
    else if (a == 0)
        return 0;
    else
        return F(a-2,level + 1) + F(a-1,level+1);

}

->F(6)
  ->F(4)
    ->F(2)
      ->F(0)
      ->F(1)
    ->F(3)
      ->F(1)
      ->F(2)
        ->F(0)
        ->F(1)
  ->F(5)
    ->F(3)
      ->F(1)
      ->F(2)
        ->F(0)
        ->F(1)
    ->F(4)
      ->F(2)
        ->F(0)
        ->F(1)
      ->F(3)
        ->F(1)
        ->F(2)
          ->F(0)
          ->F(1)