Java 基本递归方法 - 阶乘
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2802775/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Basic recursive method - factorial
提问by bob
I am practicing recursion and I can't see why this method does not seem to work. Any ideas?
我正在练习递归，我不明白为什么这种方法似乎不起作用。有任何想法吗？
public void fact()
{
fact(5);
}
public int fact(int n)
{
if(n == 1){
return 1;
}
return n * (fact(n-1));
}
}
Thanks
谢谢
采纳答案by Anthony Forloney
Your code seems to work but you are not doing anything with the returned value, put method call factor fact(5)inside of a System.out.printlnand see what you get.
您的代码似乎可以工作，但您没有对返回的值、put 方法调用fact或 afact(5)内部进行任何操作，System.out.println然后看看您得到了什么。
回答by andyczerwonka
Works fine. You're not assigning it to anything. Here's a test that'll prove it works.
工作正常。你没有将它分配给任何东西。这是一个可以证明它有效的测试。
@Test
public void testYourFactorialMethod() {
assertEquals(120, fact(5));
}
回答by Nasser Hadjloo
It is totaly wrong to write Fibonacci with recursive methods!!
用递归方法写斐波那契是完全错误的！！
It is an old famous example for how a good/bad Algorythm affect any project
这是一个很好的/坏的算法如何影响任何项目的著名例子
if you write Fibonatcci recursive, for calculating 120you need 36 year toget the result!!!!!!
如果你写斐波那契递归，计算120你需要36年才能得到结果！！！！！！
public static int Fibonacci(int x)
{ // bad fibonacci recursive code
if (x <= 1)
return 1;
return Fibonacci(x - 1) + Fibonacci(x - 2);
}
in dot net 4.0 there is a new type name BigInteger and you can use it to make a better function
在 dot net 4.0 中有一个新的类型名称 BigInteger，您可以使用它来制作更好的功能
using System; using System.Collections.Generic; using System.Numerics; //needs a ref. to this assembly
使用系统；使用 System.Collections.Generic; 使用 System.Numerics；//需要一个参考。到这个大会
namespace Fibonaci
{
public class CFibonacci
{
public static int Fibonacci(int x)
{
if (x <= 1)
return 1;
return Fibonacci(x - 1) + Fibonacci(x - 2);
}
public static IEnumerable<BigInteger> BigFib(Int64 toNumber)
{
BigInteger previous = 0;
BigInteger current = 1;
for (Int64 y = 1; y <= toNumber; y++)
{
var auxiliar = current;
current += previous;
previous = auxiliar;
yield return current;
}
}
}
}
and you can use it like
你可以像这样使用它
using System;
using System.Linq;
namespace Fibonaci
{
class Program
{
static void Main()
{
foreach (var i in CFibonacci.BigFib(10))
{
Console.WriteLine("{0}", i);
}
var num = 12000;
var fib = CFibonacci.BigFib(num).Last();
Console.WriteLine("fib({0})={1}", num, fib);
Console.WriteLine("Press a key...");
Console.ReadKey();
}
}
}
and in this case you can calculate 12000less than a second. so
在这种情况下，您可以计算12000不到一秒钟。所以
Using Recursive methos is not always a good idea
使用递归方法并不总是一个好主意
Above code imported from Vahid Nasiri blog whiche wrote in Persian
以上代码从Vahid Nasiri 博客导入，该博客用波斯语编写
回答by polygenelubricants
The recursion part is fine; you're just not using its returnvalue, which gets discarded. Here's a complete Java application of your factorial code, slightly jazzed-up for educational purposes:
递归部分很好；你只是没有使用它的return价值，它被丢弃了。这是您的阶乘代码的完整 Java 应用程序，出于教育目的而略加修饰：
public class Factorial {
public static String fact(int n) {
if(n == 1){
return "1";
}
return n + " * " + (fact(n-1)); // what happens if you switch the order?
}
public static void main(String[] args) {
System.out.println(fact(5));
// prints "5 * 4 * 3 * 2 * 1"
}
}
回答by Jonas Fagundes
A simplified version of your code:
代码的简化版本：
public int fact(int n)
{
if(n == 1){
return 1;
}
return n * (fact(n-1));
}
could be just:
可能只是：
public int fact(int n)
{
return n == 1 ? 1 : n * fact(n - 1);
}
but your code is not wrong, this is just another style (if you are not used to ternary operator keep the way it is). I prefer use the ternary operator in these cases (observe that the code is side effect free).
但是你的代码没有错，这只是另一种风格（如果你不习惯三元运算符保持原样）。在这些情况下，我更喜欢使用三元运算符（注意代码没有副作用）。
回答by user1730535
public class Recursive {
public static void main(String[] argss) {
System.out.print(fac(3));
}
public static int fac(int n) {
int value = 0;
if (n == 0) {
value = 1;
} else {
value = n * fac(n - 1);
}
return value;
}
}
// out put 6
回答by Amruth M Raj
static int factorial(int x) {
int result;
if (x == 1) {
return 1;
}
// Call the same method with argument x-1
result = factorial(x – 1) * x;
return result;
}
For complete example check this
对于完整的例子，检查这个
http://answersz.com/factorial-program-in-java-using-recursion/
http://answersz.com/factorial-program-in-java-using-recursion/
回答by Alex Mapley
Try something like this: (Or maybe try this directly)
尝试这样的事情：（或者直接尝试）
public class factorial {
private static int factorial( int n ){
if (n > 1) {
return n * (factorial(n-1));
} else {
return 1;
}
}
public static void main(String[] args) {
System.out.println(factorial(100));
}
}
