如何产生一个1000以内的斐波那契序列？

Question

正在写一网站上的Javascript挑战题：给定一个数n，编写函数返回<=n的所有斐波那契数。

Example:
    Fibonacci(4) -> 1, 1, 2, 3
    Fibonacci(5) -> 1, 1, 2, 3, 5
    Fibonacci(7) -> 1, 1, 2, 3, 5

如何产生Fibonacci数我会，只需给出思路即可，代码我自己实现。

Answer 1

既然会产生F数，那为什么不会做这一题呢？只要在for循环里判断一下有没有达到n即可，没达到就继续产生，然后把产生的F数存入一个数组，达到了就跳出循环并返回该数组

Answer 2

function fib($n)
{
    $fn_1 = 1;
    $fn_2 = 1;
    $fib = 1;
    $res = array($fn_1,$fn_2);
    while (true) 
    {
        $fn_1 = $fn_2;
        $fn_2 = $fib;

        $fib = $fn_1+$fn_2;
        if ($fib < $n) 
        {
            $res[] = $fib;
        }
        else
        {
            break;
        }
    }
    return $res;
}

$n = 15;
print_r(fib($n));
echo PHP_EOL;

Answer 3

生成小于某数的序列
def fibSequenceLessThanNum(num):
    a, b = 0, 1
    result = []
    while a < num:
        result.append(a)
        a, b = b, a + b
    return result
    
fibSeqLessThanNum(1000)
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987]
>>> 

Answer 4

"""
之前题目理解错误,修改下答案. 方法一：
"""
def fib_list_generate(n=0):
    fib_number_first, fib_number_second = 1, 1
    while (fib_number_first < n):
        yield fib_number_first
        fib_number_first, fib_number_second = fib_number_second, fib_number_first + fib_number_second


for i in fib_list_generate(1000):
    print i
    
"""
之前题目理解错误,修改下答案. 方法二：
"""
def fib_list_generate_001(n=0):
    fib_list = []
    fib_number_first, fib_number_second = 1, 1
    while(fib_number_first < n):
        fib_list.append(fib_number_first)
        fib_number_first, fib_number_second = fib_number_second, fib_number_first + fib_number_second
    return  fib_list


print fib_list_generate_001(1000)

Answer 5

一个个加上去啊，超过n就停

Answer 6

1000位以后，那数得多大呀，循环什么的效率都低了吧

Answer 7

用迭代：f(n)=f(n-1)+f(n-2)

Answer 8

function fibonacci(n) {
    var i = 1, j = 1;
    var arr = [i, j];
    var last = 1;

    while (true) {
        last = i + j;
        if(last>n)break;
        i = j;
        j = last;
        arr.push(last);
    }
    console.log(arr.join(','));
}

fibonacci(1000);//1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987

Answer 9

先说斐波那契数：

     arr[i] = arr[i-1] + arr[i-2]，

JavaScript中的循环判断是否小于n，在循环体内对i进行判断i<2的情况，后面你懂的。。

Answer 10

RTFM

Answer 11

a[i] = a[i - 1] + a[i - 2];