如何在第一次调用此方法时以及在进一步的递归调用中保留"n"的值？

Question

public void staircase(int n)
{
// I need to retain the value of "n" going forward in my recursion calls 
// Also, everything needs to be done inside the method!
int check = 0;
if(n != 0)
{
for (int i = 0;i< n; i++)
{
// method to draw
if (check < (n - 1))
{
Console.Write(" ");
check++;
}
else
Console.Write("#");
}
if (check != 0)
{
Console.Write("n");
staircase(check);
}
}
Console.ReadLine();
}

Answer 1

一种选择是向方法添加第二个参数并将其设置为某个默认值，可能null(通过使其成为可为空的int(。然后，仅当它是null时才将其设置为n的值(指示这是第一次调用该方法(，以便可以将其传递给后续递归调用：

public void staircase(int number, int? original = null)
{
// On the first iteration, 'original' should be null, so assign 'n' to it
if (original == null) original = number;
// * Other code remains the same *
// Pass 'original' to recursive calls, which should be the first value of 'n'
if (check != 0)
{
Console.Write("n");
staircase(check, original);
}       
// * Other code remains the same *
}

Answer 2

您可以使用如下本地方法：

public void StairCase(int n)
{
void StairCaseInner(int n_inner)
{
// you can still access `n` in here.

Console.WriteLine(n);

int check = 0;
if (n_inner != 0)
{
for (int i = 0; i < n_inner; i++)
{
// method to draw
if (check < (n_inner - 1))
{
Console.Write(" ");
check++;
}
else
Console.Write("#");
}
if (check != 0)
{
Console.Write("n");
StairCaseInner(check);
}
}
}
StairCaseInner(n);

Console.ReadLine();
}

当我调用StairCase(5);时，我从中得到的输出是：

5 # 5  # 5 # 5  # 5 #

我很想知道该方法对原始n有什么作用。