JavaScript 中的斐波那契数列
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Fibonacci series in JavaScript
提问by sumon
function fib(n) {
const result = [0, 1];
for (var i = 2; i <= n; i++) {
const a = (i - 1);
const b = (i - 2);
result.push(a + b);
}
return result[n];
}
console.log(fib(8));The output of the code above is 13. I don't understand the for loop part. In very first iteration i = 2, but after second iteration i = 3so a = 2and b = 1and third iteration i = 4so a = 3, b = 2, and so on... If it's going on final sequence will be :
[0, 1, 1, 3, 5, 7, 9, 11], which is incorrect. The correct sequence will be [0, 1, 1, 2, 3, 5, 8, 13]
上面代码的输出是13. 我不明白 for 循环部分。在第一次迭代中i = 2，但在第二次迭代后i = 3如此a = 2和b = 1第三次迭代i = 4如此a = 3，b = 2......如果它继续进行，最终序列将是 :
[0, 1, 1, 3, 5, 7, 9, 11]，这是不正确的。正确的顺序是[0, 1, 1, 2, 3, 5, 8, 13]
回答by Saif Ur Rahman
You were not using the previous two numbers that are already in the array to > generate the new fibonacci number to be inserted into the array.
您没有使用数组中已有的前两个数字来 > 生成要插入到数组中的新斐波那契数。
https://www.mathsisfun.com/numbers/fibonacci-sequence.html
https://www.mathsisfun.com/numbers/fibonacci-sequence.html
Here I have used the sum of result[i-2]and result[i-1]to generate the new fibonacci number and pushed it into the array.
在这里，我已经使用的总和result[i-2]并result[i-1]生成新的Fibonacci数与推到阵列。
Also to generate nnumber of terms you need the condition to be i < nand not i <= n.
此外，要生成n您需要的条件是i < n和不是的条款数i <= n。
function fib(n) {
const result = [0, 1];
for (var i = 2; i < n; i++) {
result.push(result[i-2] + result[i-1]);
}
return result; // or result[n-1] if you want to get the nth term
}
console.log(fib(8)); Return result[n-1]if you want to get the nth term.
返回result[n-1]如果你想获得的第n项。
回答by mojtaba ramezani
My solution for Fibonacci series:
我对斐波那契数列的解决方案：
const fibonacci = n =>
[...Array(n)].reduce(
(acc, val, i) => acc.concat(i > 1 ? acc[i - 1] + acc[i - 2] : i),
[]
)
回答by Cerberus
This function is incorrect. It cat be checked by just adding the console.logcall just before the function return:
此功能不正确。console.log只需在函数返回之前添加调用即可检查它：
function fib(n) {
const result = [0, 1];
for (var i = 2; i <= n; i++) {
const a = (i - 1);
const b = (i - 2);
result.push(a + b);
}
console.log(result);
return result[n];
}
console.log(fib(7));As you can see, the sequence is wrong and (for n = 7) the return value is too.
如您所见，序列是错误的，并且（对于n = 7）返回值也是错误的。
The possible change would be as following:
可能的变化如下：
function fib(n) {
const result = [0, 1];
for (var i = 2; i <= n; i++) {
const a = result[i - 1];
const b = result[i - 2];
result.push(a + b);
}
console.log(result);
return result[n];
}
console.log(fib(8));This is the "classical" Fibonacci numbers; if you really want to use the first number of 0, not 1, then you should return result[n-1], since array indexes start from zero.
这是“经典”斐波那契数列；如果你真的想使用第一个数字0，而不是1，那么你应该return result[n-1]，因为数组索引从零开始。
回答by Tom O.
One approach you could take for fibonacci sequence is recursion:
您可以对斐波那契数列采取的一种方法是递归：
var fibonacci = {
getSequenceNumber: function(n) {
//base case to end recursive calls
if (n === 0 || n === 1) {
return this.cache[n];
}
//if we already have it in the cache, use it
if (this.cache[n]) {
return this.cache[n];
}
//calculate and store in the cache for future use
else {
//since the function calls itself it's called 'recursive'
this.cache[n] = this.getSequenceNumber(n - 2) + this.getSequenceNumber(n - 1);
}
return this.cache[n];
},
cache: {
0: 0,
1: 1
}
}
//find the 7th number in the fibbonacci function
console.log(fibonacci.getSequenceNumber(7));
//see all the values we cached (preventing extra work)
console.log(fibonacci.cache);
//if you want to output the entire sequence as an array:
console.log(Object.values(fibonacci.cache));The code above is also an example of a dynamic programmingapproach. You can see that I am storing each result in a cacheobject the first time it is calculated by the getSequenceNumbermethod. This way, the second time that getSequenceNumberis asked to find a given input, it doesn't have to do any actual work - just grab the value from cacheand return it! This is an optimization technique that can be applied to functions like this where you may have to find the value of a particular input multiple times.
上面的代码也是动态编程方法的一个例子。您可以看到我将每个结果在cache第一次由该getSequenceNumber方法计算时存储在一个对象中。这样，第二次getSequenceNumber被要求查找给定的输入时，它不需要做任何实际工作——只需从中获取值cache并返回它！这是一种优化技术，可以应用于这样的函数，您可能需要多次找到特定输入的值。
回答by patki
simple solution for Fibonacci series:
斐波那契数列的简单解决方案：
function fib(n){
var arr = [];
for(var i = 0; i <n; i++ ){
if(i == 0 || i == 1){
arr.push(i);
} else {
var a = arr[i - 1];
var b = arr[i - 2];
arr.push(a + b);
}
}
return arr
}
console.log(fib(8))
回答by ShawnTG71
This is certainly one of those "more than one way to clean chicken" type situations, this JavaScript method below works for me.
这当然是“不止一种清洁鸡肉的方法”类型的情况之一，下面的这个 JavaScript 方法对我有用。
function fibCalc(n) {
var myArr = [];
for (var i = 0; i < n; i++) {
if(i < 2) {
myArr.push(i);
} else {
myArr.push(myArr[i-2] + myArr[i-1]);
}
}
return myArr;
}
fibCalc(8);
When called as above, this produces [0,1,1,2,3,5,8,13] It allows me to have a sequence of fib numbers based on n.
当如上所述调用时，这会产生 [0,1,1,2,3,5,8,13] 它允许我有一个基于 n 的 fib 数字序列。
回答by Adrian Brand
const fib = n => {
const array = Array(n);
for (i = 0; i < array.length; i++) {
if (i > 1) {
array[i] = array[i - 1] + array[i - 2];
} else {
array[i] = 1;
}
}
return array;
}
console.log(fib(5))回答by AYO O.
function fib(n) {
const result = [0];
if (n > 1) {
result.push(1);
for (var i = 2; i < n; i++) {
const a = result[result.length - 1]
const b = result[result.length - 2];
result.push(a + b);
}
}
console.log(result);
}
回答by Viraj
i came up with this solution to get the n index fibonacci value.
我想出了这个解决方案来获得 n 指数斐波那契值。
function findFac(n){
if (n===1)
{
return [0, 1];
}
else
{
var s = findFac(n - 1);
s.push(s[s.length - 1] + s[s.length - 2]);
return s;
}
}
function findFac0(n){
var vv1 = findFac(n);
return vv1[n-1];
}
console.log(findFac0(10));
回答by Kiran Racherla
Here, you have it, with few argument check, without using exception handling
在这里，您拥有它，只需很少的参数检查，无需使用异常处理
function fibonacci(limit){
if(typeof limit != "number"){return "Please enter a natural number";}
if(limit <=0){
return "limit should be at least 1";
}
else if(limit == 1){
return [0];
}
else{
var series = [0, 1];
for(var num=1; num<=limit-2; num++){
series.push(series[series.length-1]+series[series.length-2]);
}
return series;
}
}
