英文:
Stackoverflow error while calling another method within one
问题
import java.util.Scanner;
public class lab6_Perea {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("你好！欢迎使用集合计算器。请选择您想要执行的计算：\n\t1. 指数运算\n\t2. 阶乘\n\t3. 组合\n\t4. 排列");
int option = sc.nextInt();
switch (option) {
case 1:
System.out.println("请输入底数和指数：");
int x = sc.nextInt();
int y = sc.nextInt();
System.out.print("答案是：" + power(x, y));
break;
case 2:
System.out.println("请输入一个数字：");
int z = sc.nextInt();
System.out.print(z + " 的阶乘是 " + factorial(z));
break;
case 3:
System.out.println("请输入两个数字来计算组合：");
int a = sc.nextInt();
int b = sc.nextInt();
System.out.print("总组合数为：" + combination(a, b));
break;
case 4:
// FIX ME 调用组合方法
break;
}
}
public static int power(int n, int m) {
if (m == 0)
return 1;
else
return (n * power(n, m - 1));
}
public static int factorial(int n) {
if (n == 0)
return 1;
else
return (n * factorial(n - 1));
}
public static int combination(int n, int r) {
int combo = factorial(n) / (factorial(r) * factorial(n - r));
return combo;
}
}
英文:
I'm trying to compute the combination
procedure(factorial(n)/(factorial(r)*factorial(n-r)).
When I call the factorial method from within the combination method, when running I get a stackoverflow error before the calculation.
The instructions read "This method should receive two integers as its parameters, n and r. Your combination method should call the factorial method inside its definition by using it in the following formula:"
Any help will be appreciated we haven't really covered stackoverflow so I'm a little lost. Thank you!!
import java.util.Scanner;
public class lab6_Perea{
public static void main(String[] args){
Scanner sc = new Scanner (System.in);
System.out.println("Hello! Welcome to Set calculator. Please select the calculation you want to perform \n\t1. Power \n\t2. Factorial \n\t3. Combination \n\t4. Permutation");
int option = sc.nextInt();
switch(option){
case 1:
System.out.println("Please enter the base and exponent: ");
int x = sc.nextInt();
int y = sc.nextInt();
System.out.print("The answer is: " + power(x, y));
break;
case 2:
System.out.println("Please enter a number: ");
int z = sc.nextInt();
System.out.print("The facotrial of " + z + " is " + factorial(z));
break;
case 3:
System.out.println("Please enter 2 numbers to calculate the Combination: ");
int a = sc.nextInt();
int b = sc.nextInt();
System.out.print("The total combination is: " + combination(a, b));
break;
case 4:
// FIX ME call Combination method
break;
}
}
public static int power(int n, int m){
if(m==0)
return 1;
else
return (n * power(n, m-1));
}
public static int factorial(int n){
if(n == 0)
return 0;
else
return (n * factorial(n-1));
}
public static int combination ( int n, int r){
int combo =factorial(n)/(factorial(r)*factorial(n-r));
return combo;
}
}
答案1
得分: 0
以下是翻译好的内容：
使用这个函数：
public static int factorial(int n){
if(n <= 1)
return 1;
else
return (n * factorial(n-1));
}
而且你不需要测试 n 是否为正数。你所写的递归进入了堆栈溢出，因为 n-r < 0，而你原来的规则（如果 n==0）从未被遇到。
英文:
Use this function:
public static int factorial(int n){
if(n <= 1)
return 1;
else
return (n * factorial(n-1));
}
And you do not have to test n for positivity. The recursion you wrote went in StackOverflow because n-r<0 and your original rule (if n==0) was never encountered.
答案2
得分: -1
使用 return 1 if n==0
public static int factorial(int n){
if(n == 0)
return 1;
else
return (n * factorial(n-1));
}
请检查 n>=r 的条件。
希望这能够起作用！
英文:
Use return 1 if n==0
public static int factorial(int n){
if(n == 0)
return 1;
else
return (n * factorial(n-1));
}
Please check n>=r condition.
Hope, this will work!
答案3
得分: -1
因为在递归中可以使用的递归次数是有限的。
为了克服这个问题，你可以在阶乘函数内部使用一个for循环。
英文:
It's because there is some limit on the number of recursions one can use.
To overcome it, you can use a for loop inside the factorial function.
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