java - 递归问题？

Question

所以这是我的第一个递归函数（我希望！），我不确定为什么它不起作用，（又名红线）有什么想法吗？

int myFactorial(int C) { //underlined, expects ";"

    int n = Integer.parseInt(objectsChooseField.getText());
    int r = Integer.parseInt(chooseFromField.getText());

    if (C == 1){
        return 1; //underlined, cannot return value from method whose result type is void
    }
    return (C*(myFactorial(n/(r(n-r))))); //underlined
}

Answer 1

这里：r(n-r)。

r不是函数，而是局部变量int。

你的意思是r * (n - r)？

Answer 2

回避语句不应该是：

return ( C * myFactorial ( C - 1 ) );

Answer 3

你的方程没有说明任何递归。教师可以通过递归计算。

public int MyMethod() {
int n = Integer.parseInt(objectsChooseField.getText());
int r = Integer.parseInt(chooseFromField.getText());
int result = C( n, r );
}

public int C( int n, int r ) {
  int res = faculty( n ) / ( faculty( r ) * ( n - r ));
  return res;
}

//--- Computing the faculty itself could be done by recursion :-)
public int faculty( n ) {
  if ( n > 1 )
    return n * faculty( n - 1 );
  return 1;
}

Answer 4

好的。所以你有 ActionPerformed 方法。输入如下代码：

private void calculateButtonActionPerformed(java.awt.event.ActionEvent evt) {
    int n = Integer.parseInt(objectsChooseField.getText());
    int r = Integer.parseInt(chooseFromField.getText());
    int result = faculty( n ) / ( faculty( r ) * ( n - r ));
    //--- Output result to somewhere
}

以及教师计算方法本身：

/** Computes the faculty of n */
public int faculty( n ) {
  if ( n > 1 )
    return n * faculty( n - 1 );
  return 1;
}