第 3 题：多种方式实现斐波那契数列 #9

lgwebdream · 2020-06-19T11:50:39Z

欢迎在下方发表您的优质见解

Genzhen · 2020-06-22T14:14:43Z

参考实现

数学上是以递归的方法来定义

F(0) = 0;
F(1) = 1;
F(n) = F(n - 1) + F(n - 2);

公式版：递归

function fib(n) {
  if(n < 0) throw new Error('输入的数字不能小于0');
  if (n < 2) {
    return n;
  }
  return fib(n - 1) + fib(n - 2);
}

正常递归版本是一个既简单又直接的逻辑，但是这个版本有个问题就是存在大量重复计算。如：当 n 为 5 的时候要计算fib(4) + fib(3)当 n 为 4 的要计算fib(3) + fib(2) ，这时fib(3)就是重复计算了。运行 fib(50) 等半天才会出结果。

迭代:for 循环

function fib(n) {
  if(n < 0) throw new Error('输入的数字不能小于0');
  let f0 = 0,
    f1 = 1,
    curFib = f0;
  if (n < 2) {
    return n;
  }
  for (let i = 1; i < n; i++) {
    curFib = f0 + f1;
    f0 = f1;
    f1 = curFib;
  }
  return curFib;
}

这个版本没有重复计算问题，速度也明显快了很多。这并不代表循环比递归好。循环的问题在于状态变量太多，为了实现 fib 这里使用了 4 个状态变量(f0,f1,curFib,i) 而状态变量在写、修改、删除的过程中需要格外小心。状态变量多了阅读起来也不那么优美了。

去除重复计算的递归版本

function fib(n) {
  if(n < 0) throw new Error('输入的数字不能小于0');
  if (n < 2) return n;
  function _fib(n, a, b) {
    if (n === 0) return a;
    return _fib(n - 1, b, a + b);
  }
  return _fib(n, 0, 1);
}

把前两位数字做成参数巧妙的避免了重复计算，性能也有明显的提升。n 做递减运算，前两位数字做递增（斐波那契数列的递增）,这段代码一个减，一个增。

基于 ES6 Generator 实现

function* fib(n) {
  if(n < 0) throw new Error('输入的数字不能小于0');
  let f0 = 1,
    f1 = 1,
    count = 0;
  while (count < n) {
    yield f0;
    [f0, f1] = [f1, f0 + f1];
    count++;
  }
}

数组方法

function fib(n) {
  if(n < 0) throw new Error('输入的数字不能小于0');
  if (n < 2) {
    return n;
  }
  let list = [];
  list[0] = 0;
  list[1] = 1;
  for (let i = 1; i < n; i++) {
    list[i + 1] = list[i] + list[i - 1];
  }
  return list[n];
}

metal88888 · 2020-07-11T02:30:20Z

function fibonacci(n, map = {}) {
if (n == 1 || n == 2) return n;
if (map[n]) return map[n];
let data = fibonacci(n - 1, map) + fibonacci(n - 2, map)
map[n] = data
return data
}

luobolin · 2020-07-12T07:14:59Z

export const fibonacci = (n: number): number => {
    const base = Math.sqrt(5)
    return (((1 + base) / 2) ** n - ((1 - base) / 2) ** n) / base
}
console.log(fibonacci(8))

thinkfish · 2020-07-12T09:24:30Z

function fibonacci (n) {
    if (!n) {
       return 0
    }
    if(n===1 || n === 2) {
       return 1
    }
    let num1 = 1
    let num2 = 1
    let num3 = 0
    for (let i = 2; i <  n; i++){
          num3 = num1 + num2
          num1 = num2
          num2 = num3
   }
  return num3
}

时间复杂度 O(n)

thunderstromcd · 2020-07-13T07:09:14Z

使用generator

const createAry = function(args){
  
    let handler = {
        get: function(target,propKey,reciever){
            return Reflect.get(target,propKey<0 ? (+propKey + target.length) : propKey,reciever)
        },
        set :function (target,propKey,value,reciever){
            return Reflect.set(target,propKey<0 ? (+propKey + target.length) : propKey,value,reciever)
        }
    }
    return new Proxy(args,handler)
}

var ary= createAry([1,2,3,4])

console.log(ary[-1]);

ary[-1]='a'
console.log(ary[-1]);

123456zzz · 2020-07-13T14:06:59Z

const fibonacci = (n: number) => {
  let pre = 1,
    cur = 1;
  if (n === 1 || n === 2) {
    return 1;
  }
  for (let i = 3; i <= n; i++) {
    [pre, cur] = [cur, pre + cur];
  }
  return cur;
};

console.assert(fibonacci(1) === 1);

console.assert(fibonacci(2) === 1);

console.assert(fibonacci(3) === 2);

console.assert(fibonacci(4) === 3);

523451928 · 2020-07-15T07:36:19Z

function fibonacci(n) {
if (n <= 1) {
return n
}
return fibonacci(n - 1) + fibonacci(n - 2)
}
fibonacci(5)

cool-518 · 2020-07-17T06:41:42Z

@Genzhen 写的明显有问题还那么多人点赞，笑死

Genzhen · 2020-07-20T00:53:34Z

@Genzhen 写的明显有问题还那么多人点赞，笑死

@cool-518 是基于ES6 Generator实现这个？输出是没有问题的

function* fib(n) {
  if (n < 0) throw new Error("输入的数字不能小于0");
  let f0 = 1,
    f1 = 1,
    count = 0;
  while (count < n) {
    yield f0;
    [f0, f1] = [f1, f0 + f1];
    count++;
  }
}
for(let n of fib(10)){
  console.log(n);
}
// 1 1 2 3 5 8 13 21 34 55

yaooooooooo · 2020-07-20T01:17:56Z

function fabonacci(i , i1 = 1 , i2 = 1){
if(i<= 1) return i2;
return fabonacci(i -1, i2, i1 + i2)
}

GolderBrother · 2020-07-20T13:19:45Z

/**

求斐波那契数列第n项的值
@param {*} n 第n项
@param {*} map 第n项的值，缓存计算结果，防止重复计算
*/
function fibonacci(n, map = {}) {
if (n === 1 || n === 2) return n;
if (map[n]) return map[n];
// fn(n) = fn(n - 1) + fn(n - 2);
const result = fibonacii(n - 1, map) + fibonacii(n - 2, map);
map[n] = result;
return result;
}

/**

求斐波那契数列第n项的值
@param {*} n 第n项
*/
function fibonacci(n) {
let pre = 1,
cur = 1,
data;
if (n === 1 || n === 2) return n;
for (let i = 3; i <= n; i++) {
data = pre + cur;
pre = cur;
cur = data;
}
return data;
}

/**

求斐波那契数列第0-n项的集合
@param {*} n
*/
function fibonacci(n) {
let a = 1,
b = 1,
sum;
let arr = [1, 1];
for (let i = 3; i <= n; i++) {
sum = a + b;
a = b;
b = sum;
arr.push(sum);
}
return arr;
}

Z6T · 2020-07-29T07:40:52Z

function fn (n){
    if(n==0) return 0
    if(n==1) return 1
    return fn(n-2)+fn(n-1)
}

const fn = (n)=>{
    let pre = 1;
    let cur = 1;
    let data ;
    for(var i = 3;i<=n;i++){
// 和变成当前项cur,而当前项cur会变成pre
        data = pre + cur;
        pre = cur 
        cur = data
    }
    return data
}
console.log(fn(2))

chun1hao · 2020-07-31T02:50:42Z

function (n) {
if(n<2) return n
let a = 0,b = 1
for(let i=2;i<=n;i++){
[a, b] = [b, a+b]
}
return b
}

JJL-SH · 2020-08-20T10:03:42Z

// 0,1,1,2,3,5,8,13,21
const fbnq = (num) =>
  Array.from(new Array(num)).reduce((cacheArr, it, index) => {
    if (index > 1) {
      cacheArr.push(cacheArr[index - 2] + cacheArr[index - 1]);
    } else {
      cacheArr.push(index);
    }

    return cacheArr;
  }, []);

console.log(fbnq(9));

huzedong2015 · 2020-08-28T06:57:11Z

性能最低到50就会卡死

function fibonacci(n) {
    if (n < 3) {
        return 1;
    }

    return fibonacci(n - 2) + fibonacci(n - 1);
}

console.time("fibonacci1");
console.log(fibonacci(40));
console.timeEnd("fibonacci1");

性能大幅提升

function fibonacci2(n) {
    const arr = [1, 1, 2];
    const arrLen = arr.length;

    if (n <= arrLen) {
        return arr[n];
    }

    for (let i = arrLen; i < n; i++) {
        arr.push(arr[i - 1] + arr[ i - 2]);
    }

    return arr[arr.length - 1];
}

console.time("fibonacci2");
console.log(fibonacci2(10000));
console.timeEnd("fibonacci2");

性能进一步提升

function fibonacci3(n) {
    let pre1 = 1;
    let pre2 = 1;
    let current = 2;

    if (n <= 2) {
        return current;
    }

    for (let i = 2; i < n; i++) {
        pre1 = pre2;
        pre2 = current;
        current = pre1 + pre2;
    }

    return current;
}

console.time("fibonacci2");
console.log(fibonacci3(10000));
console.timeEnd("fibonacci2");

RainyNight9 · 2020-09-08T06:26:44Z

function fabonacci(n) {
  let arr = [0, 1];
  for (let i = 2; i <= n; i++) {
      arr.push(arr[i-2] + arr[i-1]);
  }
  return arr;
}

ghost · 2020-09-08T11:56:58Z

function fabonacci(n, n_1, n_2) {
if(n==1)
return n_1;
return fabonacci(n-1, n_1+n_2, n_1);
}

carbrokers · 2020-09-10T08:45:39Z

function fabonacci(max) {
  if (max === 1 || max === 2) {
    return 1;
  }
  return fabonacci(max - 1) + fabonacci(max - 2);
}

NameWjp · 2020-10-29T02:57:30Z

function fibonacci(n) {
  if (n === 1) {
    return 1
  } else if (n === 2) {
    return 1
  } else {
    return fibonacci(n -1) + fibonacci(n - 2)
  }
}

ruandao · 2020-10-29T04:00:16Z


const fibonacci = (n) => {
    if(n < 3) {
        return 1
    }
    const arr = [1, 1];
    for(let x=3;x<=n;x++) {
        arr.push(arr[x-2] + arr[x-3])
   }
   return arr[n-1];
}

chengazhen · 2020-11-03T10:00:41Z

  function fun(length = 10, a = 0, b = 1) {
      let arr = []
      do {
        if (arr.length < 2) {
          arr.push(a)
          arr.push(b)
        } else {
          arr.push(arr[arr.length - 2] + arr[arr.length - 1])
        }
      } while (arr.length < length);
      return arr
    }
    
    console.log(fun(20));
// [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

markduan · 2020-11-12T04:54:57Z

function fibonacci(n) {
  if (n < 2) return n

  let a = BigInt(0), b = BigInt(1)

  for (let i = 2; i <= n; i++) {
    [a, b] = [b, a + b]
  }

  return b
}

Jam-Chu · 2021-02-19T12:47:49Z

    // =============无备忘录===================
    function fibonacci(n) {
      if (n === 1 || n === 2) { return 1 };
      return fibonacci(n - 1) + fibonacci(n - 2);
    }
    console.log(fibonacci(30))


    // =============有备忘录===================
    function fibonacci_memo(n, memo = {}) {
      if (n === 1 || n === 2) return 1;
      if (memo[n]) return memo[n];
      let data = fibonacci_memo(n - 1, memo) + fibonacci_memo(n - 2, memo);
      memo[n] = data;
      return data;
    }
    console.log(fibonacci_memo(30))
    // =============for迭代===================
    function fibonacci_for(n) {
      if (n === 1 || n === 2) return 1;
      let total = 0, cur = 1, pre = 1;
      for (let i = 3; i <= n; i++) {
        total = (pre + cur);
        pre = cur;
        cur = total;
      }
      return total;
    }
    console.log(fibonacci_for(30))
    // =============generater步进===================
    function fibonacci_gen(n) {
      function* gen() {
        let total = 0, cur = 1, pre = 1
        while (1) {
          total = cur + pre;
          yield total;
          [cur, pre] = [total, cur];
        }
      }
      let iterator = gen();
      let data = 0;
      if (n === 1 || n === 2) data = 1;
      else {
        for (i = 3; i <= n-1; i++) {
          iterator.next();
        }
        data = iterator.next().value
      }
      return data;
    }
    console.log(fibonacci_gen(30))

Jam-Chu · 2021-02-19T12:47:59Z

    // =============无备忘录===================
    function fibonacci(n) {
      if (n === 1 || n === 2) { return 1 };
      return fibonacci(n - 1) + fibonacci(n - 2);
    }
    console.log(fibonacci(30))


    // =============有备忘录===================
    function fibonacci_memo(n, memo = {}) {
      if (n === 1 || n === 2) return 1;
      if (memo[n]) return memo[n];
      let data = fibonacci_memo(n - 1, memo) + fibonacci_memo(n - 2, memo);
      memo[n] = data;
      return data;
    }
    console.log(fibonacci_memo(30))
    // =============for迭代===================
    function fibonacci_for(n) {
      if (n === 1 || n === 2) return 1;
      let total = 0, cur = 1, pre = 1;
      for (let i = 3; i <= n; i++) {
        total = (pre + cur);
        pre = cur;
        cur = total;
      }
      return total;
    }
    console.log(fibonacci_for(30))
    // =============generater步进===================
    function fibonacci_gen(n) {
      function* gen() {
        let total = 0, cur = 1, pre = 1
        while (1) {
          total = cur + pre;
          yield total;
          [cur, pre] = [total, cur];
        }
      }
      let iterator = gen();
      let data = 0;
      if (n === 1 || n === 2) data = 1;
      else {
        for (i = 3; i <= n-1; i++) {
          iterator.next();
        }
        data = iterator.next().value
      }
      return data;
    }
    console.log(fibonacci_gen(30))

chessyu · 2021-03-01T09:03:01Z

function fibonacci(num){
 var arr = [];
 for(var i = 0; i<num; i++){
     if(i<2){
        arr.push(1);
     }else{
        arr.push(arr[i-2] + arr[i-1])
     }
 }   
 return arr;
}
fibonacci(10)
//[1, 1, 2, 3, 5, 8, 13, 21, 34, 55]

DakerHub · 2021-05-02T10:18:04Z

使用generator

const createAry = function(args){
  
    let handler = {
        get: function(target,propKey,reciever){
            return Reflect.get(target,propKey<0 ? (+propKey + target.length) : propKey,reciever)
        },
        set :function (target,propKey,value,reciever){
            return Reflect.set(target,propKey<0 ? (+propKey + target.length) : propKey,value,reciever)
        }
    }
    return new Proxy(args,handler)
}

var ary= createAry([1,2,3,4])

console.log(ary[-1]);

ary[-1]='a'
console.log(ary[-1]);

答错题了兄弟

qq820137708 · 2021-05-27T07:13:41Z

// 尾递归
function Fibonacci(n, a= 1, b= 1) {
    if(n <= 1) {return b};
    return Fibonacci(n - 1, b, a + b);
}
console.log(Fibonacci(10))

MR-YangXu · 2021-05-31T13:56:51Z

欢迎在世界发表您的优质观点
// 这是输出对应的索引值
function flat(n) {
if (n < 0) throw new Error('输入的数字不能小于0');
return Array.from({length: n + 1}).reduce((t, v, i) => (i > 1 && t.push(t[i - 1] + t[i - 2]), t), [0, 1])[n]
}
// 这是输出对应长度的数组
function flat(len = 2) {
return Array.from({length: len}).reduce((t, v, i) => (i > 1 && t.push(t[i - 1] + t[i - 2]), t), [0, 1])
}

Luoyuda · 2021-06-09T04:20:11Z

    var fib = (n) => {
        if(n <= 1) return n
        return fib(n - 1) + fib(n - 2)
    }
    console.time('fib-1')
    console.log(fib(20))
    console.timeEnd('fib-1')
    var fib = (n) => {
        if(n <= 1) return n
        fib.cache = fib.cache || {}
        if(fib.cache[n]) return fib.cache[n]
        fib.cache[n] = fib(n - 1) + fib(n - 2)
        return fib.cache[n]
    }
    console.time('fib-2')
    console.log(fib(20))
    console.timeEnd('fib-2')
    var fib = (n) => {
        if(n <= 1) return n
        let one = 0
        let two = 1
        for(let i = 2; i <= n; i++){
            let o = one + two
            one = two
            two = o
        }
        return two
    }
    console.time('fib-3')
    console.log(fib(20))
    console.timeEnd('fib-3')

geekftz · 2021-06-18T15:41:46Z

// 记录fib(n - 1)和fib(n-2) 避免重复计算从下标2开始向上计算
function fib1(n) {
if (n < 0) throw new Error('输入的数字不能小于0');

if (n < 2) {
return n;
}

let f0 = 0;
let f1 = 1;
let fn = f0;

// 从下标为2开始
let i = 2;

while (i <= n) {
fn = f0 + f1;
f0 = f1;
f1 = fn;
i++;
}

return fn;
}

var n = 3;
var res1 = fib1(n);
console.log('%c res1 = %s', 'color: #007300', res1);

Flysky12138 · 2021-07-01T16:10:58Z

function fib(n) {
    let [cur, acc] = [1, 1]
    for (let i = 2; i < n; i++) [cur, acc] = [acc, acc + cur]
    return acc
}
console.log(fib(8))  // 21

xuxichen · 2021-07-12T09:54:26Z

function fibonacci(length) {
    var res = []
    var n = 0
    for (let i = 0; i < length; i++) {
        res.push(n)
        if(i==0) {
            n=1
        } else {
            n=n+res[i-1]
        }
    }
    return res
}

console.log(fibonacci(10));

Lincoln-w · 2021-09-16T09:35:05Z

我打印出一个数组
function fibonaci(n){
if(n<1){
return "err"
}
const myarray = new Array()
for (let i = 0;i<n;i++) {
if(i<=1){
myarray.push(i)
continue
}else{
let newn = myarray[i-1] + myarray[i-2]
myarray.push(newn)
}
}
return myarray
}

		// 调用斐波那契方法
		const newFibonaci = fibonaci(10)
		console.log(newFibonaci)

Hunterang · 2021-09-29T09:45:38Z

这个还算稍微简单

funciton fb(n) {
  if(n == 1) return 1
  if(n == 2) return 1
 return fb(n-1) + fb(n-2)
}

funciton fb1(n) {
  if(n == 1) return 1
  if(n == 2) return 1
 let start=1, start1 = 1,next
 for(let i = 3; i<= n;i++) {
   next = start + start1
  start = start1
  start1 = next
 }
return next
}

nhyu · 2021-10-03T03:15:29Z

// 方法一：循环
function fib(n) {
    let val = [0, 1];
    if (n < 1) throw Error('参数n不能小于1');
    if (n < 3) return val[n - 1];

    for (let i = 2; i < n; i++) {
        const [a, b] = val;
        val = [b, a + b];
    }
    return val[1];
}
// 方法二：递归
function fib(n = 0, arr = [0, 1]) {
    if (n < 1) throw Error('参数n不能小于1');
    if (n < 3) return arr[n - 1];
    const [a, b] = arr;
    return fib(--n, [b, a + b]);
}
// 测试
console.log(fib(1)) // 0
console.log(fib(5)) // 3
console.log(fib(10)); // 34
// 经测试两种方法并无明显性能差别，用很大的数测试都瞬间出结果

zhaoxm469 · 2021-12-11T14:36:09Z

let map = new Map();
const fbnq = (n) => {
if (map.get(n)) return map.get(n);
console.count();
if (n === 0 || n === 1) return 1;
let result = fbnq(n - 1) + fbnq(n - 2);
map.set(n, result)
return result
}

console.log(fbnq(10));

AAA611 · 2022-08-25T08:44:05Z

3种方式对比

    // 1.递归
    // 优点：简单、易于理解
    // 缺点：存在大量重复计算
    function f(n) {
      if (n <= 2) return 1
      return f(n - 1) + f(n - 2)
    }
    // 2.动态规划-1
    // 优点：避免了大量的重复计算
    // 缺点：需要一个长度为 N 的数组空间，但其实我们只需要存储当前项的前两个值就行，所以空间可以优化到 O(1)
    function f1(n) {
      if (n <= 2) return 1
      let dp = [1, 1]
      for (let i = 2; i < n; i++) {
        dp[i] = dp[i - 1] + dp[i - 2]
      }
      return dp[n - 1]
    }
    // 3.动态规划-2
    // 优点：无重复计算，只需要常数个变量储存空间，空间复杂度为 O(1)
    function f2(n) {
      if (n <= 2) return 1
      let p = 1
      let q = 1
      let res
      for (let i = 2; i < n; i++) {
        res = p + q
        p = q
        q = res
      }
      return res
    }

lang711 · 2022-08-25T14:49:14Z

    function fib(index) {
      let cur = 1, next = 0, pre
      while (index-- > 0) {
        pre = next
        next = cur + next
        cur = pre
      }
      return next
    }
    //测试
    for (let i = 0; i < 10; i++) {
      console.log(fib(i));
    }

lgwebdream added 算法 teach_tag 腾讯 company labels Jun 19, 2020

Genzhen changed the title ~~第 3 题：斐波那契数列~~ 第 3 题：多种方式实现斐波那契数列 Jun 28, 2020

Genzhen closed this as completed Jul 20, 2020

Genzhen reopened this Jul 29, 2020

第 3 题：多种方式实现斐波那契数列 #9

第 3 题：多种方式实现斐波那契数列 #9

Comments

lgwebdream commented Jun 19, 2020 • edited by Genzhen

Genzhen commented Jun 22, 2020 • edited

参考实现

metal88888 commented Jul 11, 2020

luobolin commented Jul 12, 2020

thinkfish commented Jul 12, 2020 • edited

thunderstromcd commented Jul 13, 2020

123456zzz commented Jul 13, 2020

523451928 commented Jul 15, 2020

cool-518 commented Jul 17, 2020

Genzhen commented Jul 20, 2020 • edited

yaooooooooo commented Jul 20, 2020

GolderBrother commented Jul 20, 2020

Z6T commented Jul 29, 2020 • edited

chun1hao commented Jul 31, 2020

JJL-SH commented Aug 20, 2020 • edited

huzedong2015 commented Aug 28, 2020 • edited

性能最低到50就会卡死

性能大幅提升

性能进一步提升

RainyNight9 commented Sep 8, 2020

ghost commented Sep 8, 2020

carbrokers commented Sep 10, 2020

NameWjp commented Oct 29, 2020

ruandao commented Oct 29, 2020 • edited

chengazhen commented Nov 3, 2020

markduan commented Nov 12, 2020 • edited

Jam-Chu commented Feb 19, 2021

Jam-Chu commented Feb 19, 2021

chessyu commented Mar 1, 2021

DakerHub commented May 2, 2021

qq820137708 commented May 27, 2021 • edited

MR-YangXu commented May 31, 2021

Luoyuda commented Jun 9, 2021

geekftz commented Jun 18, 2021

Flysky12138 commented Jul 1, 2021

xuxichen commented Jul 12, 2021

Lincoln-w commented Sep 16, 2021

Hunterang commented Sep 29, 2021

这个还算稍微简单

nhyu commented Oct 3, 2021

zhaoxm469 commented Dec 11, 2021

AAA611 commented Aug 25, 2022 • edited

3种方式对比

lang711 commented Aug 25, 2022

lgwebdream commented Jun 19, 2020 •

edited by Genzhen

Genzhen commented Jun 22, 2020 •

edited

thinkfish commented Jul 12, 2020 •

edited

Genzhen commented Jul 20, 2020 •

edited

Z6T commented Jul 29, 2020 •

edited

JJL-SH commented Aug 20, 2020 •

edited

huzedong2015 commented Aug 28, 2020 •

edited

ruandao commented Oct 29, 2020 •

edited

markduan commented Nov 12, 2020 •

edited

qq820137708 commented May 27, 2021 •

edited

AAA611 commented Aug 25, 2022 •

edited