var x = 0;
var y = 1;
var z;
fib[0] = 0;
fib[1] = 1;
for (i = 2; i <= 10; i++) {
alert(x + y);
fib[i] = x + y;
x = y;
z = y;
}
我尝试生成一个简单的斐波那契数列，但没有输出。
请问有人能告诉我出了什么问题吗？
生成斐波那契数列
48
- methuselah
52个回答
2
sparkida，发现你的方法有一个问题。如果你检查第10个位置，它会返回54并导致所有后续值不正确。你可以在这里看到它的表现：http://jsfiddle.net/createanaccount/cdrgyzdz/5/
(function() {
function fib(n) {
var root5 = Math.sqrt(5);
var val1 = (1 + root5) / 2;
var val2 = 1 - val1;
var value = (Math.pow(val1, n) - Math.pow(val2, n)) / root5;
return Math.floor(value + 0.5);
}
for (var i = 0; i < 100; i++) {
document.getElementById("sequence").innerHTML += (0 < i ? ", " : "") + fib(i);
}
}());<div id="sequence">
</div>- geeves
1
@sparkida - 更仔细地看了一下，这个方法在第76次迭代后就完全失控了。它给出了一个值为3416454622906706的结果，但应该是3416454622906707。然后在之后的几次迭代中（第81或82次），所有的值都能被100、1000等整除。 - geeves
2
我只想用ES6编写一个尾调用优化版本来做出贡献。这很简单；
var fibonacci = (n, f = 0, s = 1) => n === 0 ? f : fibonacci(--n, s, f + s);
console.log(fibonacci(12));- Redu
1
请直接参考sqram的答案。 - vsync
2
以下是使用递归、生成器和reduce编写斐波那契数列的示例。
'use strict'
//------------- using recursion ------------
function fibonacciRecursion(n) {
return (n < 2) ? n : fibonacciRecursion(n - 2) + fibonacciRecursion(n - 1)
}
// usage
for (let i = 0; i < 10; i++) {
console.log(fibonacciRecursion(i))
}
//-------------- using generator -----------------
function* fibonacciGenerator() {
let a = 1,
b = 0
while (true) {
yield b;
[a, b] = [b, a + b]
}
}
// usage
const gen = fibonacciGenerator()
for (let i = 0; i < 10; i++) {
console.log(gen.next().value)
}
//------------- using reduce ---------------------
function fibonacciReduce(n) {
return new Array(n).fill(0)
.reduce((prev, curr) => ([prev[0], prev[1]] = [prev[1], prev[0] + prev[1]], prev), [0, 1])[0]
}
// usage
for (let i = 0; i < 10; i++) {
console.log(fibonacciReduce(i))
}- Vlad Bezden
1
这个脚本将以数字作为参数，用于指定你想让斐波那契数列到达的位置。
function calculateFib(num) {
var fibArray = [];
var counter = 0;
if (fibArray.length == 0) {
fibArray.push(
counter
);
counter++
};
fibArray.push(fibArray[fibArray.length - 1] + counter);
do {
var lastIndex = fibArray[fibArray.length - 1];
var snLastIndex = fibArray[fibArray.length - 2];
if (lastIndex + snLastIndex < num) {
fibArray.push(lastIndex + snLastIndex);
}
} while (lastIndex + snLastIndex < num);
return fibArray;
};
- Sehul Viras
1
这是我想出来的。
//fibonacci numbers
//0,1,1,2,3,5,8,13,21,34,55,89
//print out the first ten fibonacci numbers
'use strict';
function printFobonacciNumbers(n) {
var firstNumber = 0,
secondNumber = 1,
fibNumbers = [];
if (n <= 0) {
return fibNumbers;
}
if (n === 1) {
return fibNumbers.push(firstNumber);
}
//if we are here,we should have at least two numbers in the array
fibNumbers[0] = firstNumber;
fibNumbers[1] = secondNumber;
for (var i = 2; i <= n; i++) {
fibNumbers[i] = fibNumbers[(i - 1)] + fibNumbers[(i - 2)];
}
return fibNumbers;
}
var result = printFobonacciNumbers(10);
if (result) {
for (var i = 0; i < result.length; i++) {
console.log(result[i]);
}
}
- randow_user_1234
1
初学者，不太优雅，但展示了JavaScript的基本步骤和推断。
/* Array Four Million Numbers */
var j = [];
var x = [1,2];
var even = [];
for (var i = 1;i<4000001;i++){
j.push(i);
}
// Array Even Million
i = 1;
while (i<4000001){
var k = j[i] + j[i-1];
j[i + 1] = k;
if (k < 4000001){
x.push(k);
}
i++;
}
var total = 0;
for (w in x){
if (x[w] %2 === 0){
even.push(x[w]);
}
}
for (num in even){
total += even[num];
}
console.log(x);
console.log(even);
console.log(total);
- Enogwe Victor
1
斐波那契数列（一行代码实现）
斐波那契数列（递归）
斐波那契数列（迭代实现）
斐波那契数列（带有尾调用优化）
function fibonacci(n) {
return (n <= 1) ? n : fibonacci(n - 1) + fibonacci(n - 2);
}
斐波那契数列（递归）
function fibonacci(number) {
// n <= 1
if (number <= 0) {
return n;
} else {
// f(n) = f(n-1) + f(n-2)
return fibonacci(number - 1) + fibonacci(number - 2);
}
};
console.log('f(14) = ' + fibonacci(14)); // 377 function fibonacci(number) {
// n < 2
if (number <= 0) {
return number ;
} else {
var n = 2; // n = 2
var fn_1 = 0; // f(n-2), if n=2
var fn_2 = 1; // f(n-1), if n=2
// n >= 2
while (n <= number) {
var aa = fn_2; // f(n-1)
var fn = fn_1 + fn_2; // f(n)
// Preparation for next loop
fn_1 = aa;
fn_2 = fn;
n++;
}
return fn_2;
}
};
console.log('f(14) = ' + fibonacci(14)); // 377function fibonacci(number) {
if (number <= 1) {
return number;
}
function recursion(length, originalLength, previous, next) {
if (length === originalLength)
return previous + next;
return recursion(length + 1, originalLength, next, previous + next);
}
return recursion(1, number - 1, 0, 1);
}
console.log(`f(14) = ${fibonacci(14)}`); // 377- Benny Code
1
3递归实现斐波那契数列具有指数级别的复杂度，不应用于计算大的斐波那契数。 - yunandtidus
1
// using recursive approach and in one line
const fib = x => (x <= 1)? x : fib (x - 1) + fib(x -2);
fib(15); // 610
// display the 15 first
Array.from({ length: 15 }, (v, i) => fib(i));
// [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]
// using memoization approach
function fibonacci(num, memo) {
memo = memo || {};
if (memo[num]) return memo[num];
if (num === 0) return 0;
if (num === 1) return 1;
return memo[num] = fibonacci(num - 1, memo) + fibonacci(num - 2, memo);
}
fibonacci(15); // 610
// display the 15 first
Array.from({ length: 15 }, (v, i) => fibonacci(i));
// [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]
- Leo Lanese
1
我的看法：
function fibonacci(num) {
return Array.apply(null, Array(num)).reduce(function(acc, curr, idx) {
return idx > 2 ? acc.concat(acc[idx-1] + acc[idx-2]) : acc;
}, [0, 1, 1]);
}
console.log(fibonacci(10));- Jonatas Walker
1
我尝试过这个：它可能有效：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Fibonacci</title>
<style>
* {
outline: 0px;
margin: 0px;
}
input[type="number"] {
color: blue;
border: 2px solid black;
width: 99.58vw;
}
</style>
</head>
<body>
<div id="myDiv" style="color: white;background-color: blue;">Numbers Here</div>
<input type="number" id="input1" oninput="fibonacciProgram(this.value)" placeholder="Type Some Numbers Here">
<script>
function fibonacciProgram(numberCount) {
let resultElement = document.getElementById("myDiv");
resultElement.innerHTML = " ";
if (isNaN(numberCount) || numberCount <= 0) {
resultElement.innerHTML = "please enter a number";
return;
}
let firstBox = 0;
let secondBox = 1;
let swichBox;
let entries = [];
entries.push(secondBox);
while (numberCount > 1) {
swichBox = firstBox + secondBox;
entries.push(swichBox);
firstBox = secondBox;
secondBox = swichBox;
numberCount--;
}
resultElement.innerHTML = entries.join(', ');
}
</script>
</body>
</html>
- Green
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