Java中使用for语句的斐波那契数列
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Fibonacci sequence in Java using for statements
提问by Carlo Solomon Fernandez
I tried making a Java program executing the Fibonacci sequence.
我尝试制作一个执行斐波那契数列的 Java 程序。
Here's my code:
这是我的代码：
import java.io.*;
public class Fibonacci{
public static void main(String[]args){
BufferedReader Data=new BufferedReader (new InputStreamReader(System.in));
int ctr1=0;
int ctr2=0;
int num1=0;
int num2=0;
int num3=0;
try{
System.out.println("How many numbers would you want to see?");
ctr2=Integer.parseInt(Data.readLine());
for(int ans=0; ctr1==ctr2; ctr1++){
num1++;
System.out.println(num2 + "\n" + num1);
ans=num1+num2;
System.out.println(ans);
ans=num3;
}
}catch(IOException err){
System.out.println("Error!" + err);
}catch(NumberFormatException err){
System.out.println("Invald Input!");
}
}
}
Obviously, I'm a beginner in Java and I don't know how to properly use the for statement. Would somebody be kind enough to make my code work? Or maybe make a way shorter code that works. I'm a beginner so be cool. Thanks :)
显然，我是 Java 初学者，不知道如何正确使用 for 语句。有人会好心让我的代码工作吗？或者也许可以制作一种更短的代码。我是初学者，所以要冷静。谢谢 ：）
回答by Juvanis
Look at this code snippet which is much easier than yours to understand. Solution tip is simple, you keep 2 pointers for the first 2 fibonacci numbers and update them appropriately in the loop. In the example below, the loop executes 10 times, you can modify it as desired.
看看这个比你更容易理解的代码片段。解决方案提示很简单，您为前 2 个斐波那契数保留 2 个指针，并在循环中适当地更新它们。在下面的示例中，循环执行 10 次，您可以根据需要对其进行修改。
static void fibonacci() {
int ptr1 = 1, ptr2 = 1;
int temp = 0;
System.out.print(ptr1 + " " + ptr2 + " ");
for (int i = 0; i < 10; i++) {
System.out.print(ptr1 + ptr2 + " ");
temp = ptr1;
ptr1 = ptr2;
ptr2 = temp + ptr2;
}
}
Output:
输出：
1 1 2 3 5 8 13 21 34 55 89 144
1 1 2 3 5 8 13 21 34 55 89 144
回答by Joban Dhillon
Expanding on the answers, if you want to look really cool use recursion.
扩展答案，如果您想看起来很酷，请使用递归。
public class Fibonacci {
public static long fib(int n) {
if (n <= 1) return n;
else return fib(n-1) + fib(n-2);
}
public static void main(String[] args) {
int N = 300; // how many numbers you want to generate
for (int i = 1; i <= N; i++)
System.out.println(i + ": " + fib(i));
}
}
Here is Google search of what it is, hope those resources help: http://bit.ly/1cWxhUS
这是谷歌搜索它是什么，希望这些资源有帮助：http: //bit.ly/1cWxhUS
回答by Joshua Dugger
import java.util.*;
public class sequence1
{
public static void main(String[] args)
{
sequence1 fs=new sequence1();
fs.fibonacci();
}
public void fibonacci()
{
int numb1 = 1;
int numb2 = 1;
int temp = 0;
@SuppressWarnings("resource")
Scanner input=new Scanner(System.in);
System.out.println("How Many Terms? (Up To 45)");
int x=input.nextInt();
x=x-2;
System.out.println(numb1);
System.out.println(numb2);
for (int i = 0; i < x; i++)
{
System.out.println(numb1 + numb2 + " ");
temp = numb1;
numb1 = numb2;
numb2 = temp + numb2;
}
}
}
回答by loknath
This function return the fibonacci series
此函数返回斐波那契数列
/**
* @param startElement
* @param secondElent
* @param length :length of fibonacci series
* @return fibonacciseries : contain the series of fibonacci series
*/
public int[] createFibonacciSeries(int startElement, int secondElent,
int length) {
int fibonacciSeries[] = new int[length];
fibonacciSeries[0] = startElement;
fibonacciSeries[1] = secondElent;
for (int i = 2; i < length; i++) {
fibonacciSeries[i] = fibonacciSeries[i - 1]
+ fibonacciSeries[i - 2];
}
return fibonacciSeries;
}
回答by Gopi Kiran S
import java.util.*;
class MyFibonacci {
public static void main(String a[]){
int febCount = 15;
int[] feb = new int[febCount];
feb[0] = 0;
feb[1] = 1;
for(int i=2; i < febCount; i++){
feb[i] = feb[i-1] + feb[i-2];
}
for(int i=0; i< febCount; i++){
System.out.print(feb[i] + " ");
}
}
}
回答by Immers Cherub
public class FibonacciExercitiu {
public static void main(String[] args) {
int result = fib(6); //here we test the code. Scanner can be implemented.
System.out.println(result);
}
public static int fib(int n) {
int x = 1;
int y = 1;
int z = 1; //this line is only for declaring z as a variable. the real assignment for z is in the for loop.
for (int i = 0; i < n - 2; i++) {
z = x + y;
x = y;
y = z;
}
return z;
}
/*
1. F(0) = 1 (x)
2. F(1) = 1.(y) =>Becomes x for point4
3.(z)F(2) = 2 (z) =>Becomes Y for point4 // becomes X for point 5
4.(z)F(3) = 3 // becomes y for point 5
5.(z)F(4) = 5 ..and so on
*/
}
}
回答by Rafael Amsili
public static int[] fibonachiSeq(int n)
{
if (n < 0)
return null;
int[] F = new int[n+1];
F[0] = 0;
if (n == 0)
return F;
F[1] = 1;
for (int i = 2; i <= n; i++)
{
F[i] = F[i-1] + F[i-2];
}
return F;
}
回答by vinay kumar gupta
Using while loop
使用while循环
class Feb
{
static void Main(string[] args)
{
int fn = 0;
int sn = 1;
int tn = 1;
Console.WriteLine(fn);
Console.WriteLine(sn);
while (true)
{
tn = fn + sn;
if (tn >10)
{
break;
}
Console.WriteLine(tn);
fn = sn;
sn = tn;
}
Console.Read();
}
}
回答by Roqaia al rofou
public class Febonacci {
公共课费波那契{
public static void main(String[] args) {
int first =0;
int secend =1;
System.out.print(first+","+secend);
for (int k=1;k<7;k++){
System.out.print(","+(first+secend ));
if(k%2!=0)
first+=secend;
else
secend+=first;
}
}
}
回答by Jacob Bickley
I'm a beginner in java as well however I've found an easy way to create a Fibonacci number using an array. The basic principle of a Fibonacci number is the addition of the current number and the number that came before. Here is my code:
我也是 Java 的初学者，但是我找到了一种使用数组创建斐波那契数的简单方法。斐波那契数的基本原理是当前数与前数相加。 这是我的代码：
//Creation of array
int [ ] fib = new int[size];
//Assigning values to the first and second indexes of array named "fib"
fib [0] = 0;
fib [1] = 1;
//Creating variable "a" to use in for loop
int a = 1
//For loop which creates a Fibonacci number
for( int i = 2; i < size ; i++)
{
fib[i] = a;
a = fib[i] + fib[i-1];
}
