Java：检查数字是否属于斐波那契数列
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Java: check if number belongs to Fibonacci sequence
提问by helena.pimentel
I'm supposed to write a code which checks if a given number belongs to the Fibonacci sequence. After a few hours of hard work this is what i came up with:
我应该编写一个代码来检查给定的数字是否属于斐波那契数列。经过几个小时的努力，这就是我想出的：
public class TP2 {
/**
* @param args
*/
public static boolean ehFibonacci(int n) {
int fib1 = 0;
int fib2 = 1;
do {
int saveFib1 = fib1;
fib1 = fib2;
fib2 = saveFib1 + fib2;
}
while (fib2 <= n);
if (fib2 == n)
return true;
else
return false;
}
public static void main(String[] args) {
int n = 8;
System.out.println(ehFibonacci(n));
}
}
I must be doing something wrong, because it always returns "false". Any tips on how to fix this?
我一定是做错了什么，因为它总是返回“false”。有关如何解决此问题的任何提示？
回答by John3136
You continue the loop while fib2 <= n, so when you are out of the loop, fib2 is always > n, and so it returns false.
您继续循环 while fib2 <= n，因此当您退出循环时， fib2 始终为> n，因此它返回false。
回答by jakub.petr
/**
* @param args
*/
public static boolean ehFibonacci(int n) {
int fib1 = 0;
int fib2 = 1;
do {
int saveFib1 = fib1;
fib1 = fib2;
fib2 = saveFib1 + fib2;
}
while (fib2 < n);
if (fib2 == n)
return true;
else
return false;
}
public static void main(String[] args) {
int n = 5;
System.out.println(ehFibonacci(n));
}
回答by Nithya
This works. I am not sure about efficiency..but this is a foolproof program,
这有效。我不确定效率……但这是一个万无一失的程序，
public class isANumberFibonacci {
public static int fibonacci(int seriesLength) {
if (seriesLength == 1 || seriesLength == 2) {
return 1;
} else {
return fibonacci(seriesLength - 1) + fibonacci(seriesLength - 2);
}
}
public static void main(String args[]) {
int number = 4101;
int i = 1;
while (i > 0) {
int fibnumber = fibonacci(i);
if (fibnumber != number) {
if (fibnumber > number) {
System.out.println("Not fib");
break;
} else {
i++;
}
} else {
System.out.println("The number is fibonacci");
break;
}
}
}
}
}
