第 99 题：编程算法题 #153

yygmind · 2019-07-04T15:40:25Z

用 JavaScript 写一个函数，输入 int 型，返回整数逆序后的字符串。如：输入整型 1234，返回字符串“4321”。要求必须使用递归函数调用，不能用全局变量，输入函数必须只有一个参数传入，必须返回字符串。

zh1989cool · 2019-07-04T16:39:12Z

function fun(num){
    let num1 = num / 10;
    let num2 = num % 10;
    if(num1<1){
        return num;
    }else{
        num1 = Math.floor(num1)
        return `${num2}${fun(num1)}`
    }
}
var a = fun(12345)
console.log(a)
console.log(typeof a)

dorseysen · 2019-07-05T01:22:56Z

其实这种直接使用JavaScript的API更简单直观。

//  the first solution by recursion

const numToReverseStr = num => {

    if( 'number' !== typeof num ) throw '输入需为int型整数';

    if(!Math.floor(num / 10)) return num.toString();

    return (num % 10).toString() + numToReverseStr( Math.floor(num / 10) );
}

console.log(numToReverseStr(2169362));

//  the second solution not by recursion but JavaScript API

const numToReverseStr_0 = num => {

    return num.toString().split('').reverse().join('');
}

console.log(numToReverseStr_0(2169362));

Gentlemancj · 2019-07-05T01:30:15Z

`function test(num) {
var str = num + "";
if(str.length > 1) {
var newStr = str.substring(str.length - 1);
var oldStr = str.substring(0, str.length - 1);
return newStr + test(oldStr)
} else {
return num
}

}

var s = test(1234);

console.log(s);`

lhj767382286 · 2019-07-05T01:45:11Z

function test(num){
	num = typeof num === 'number' ? num.toString() : num
	if(num.length <= 1) return num
	else return num[num.length - 1] + test(num.substring(0, num.length - 1))
}

TH-coder · 2019-07-05T01:45:48Z

function numberReverse(num) {
    const str = num.toString()
    return str.length === 1 ? str : numberReverse(str.substring(1)) + str.substring(0, 1)
}

这样行不,有错望指出

huangxiaochang · 2019-07-05T01:52:33Z

function reverseNumber (num) {
if ((num / 10 >> 0) === 0) {
return (‘’ + num)
}
return ('' + num % 10 + reverseNumber(num / 10 >> 0) )
}
或者简化成三行代码
function reverseNumber (num) {
return (num / 10 >> 0) === 0 ? ('' + num) : ('' + num % 10 + reverseNumber(num / 10 >> 0) )
}

kungithub · 2019-07-05T02:01:33Z

function out(value){
  
 if(value.length===0) return '';
  
 value =  Array.isArray(value)?value:value.toString().split('');
 
  return value.pop() + out(value);

}

mcDullLi · 2019-07-05T02:07:44Z

function reverseInt(intNumber) {
if (!(intNumber%10)) {
return ''
}
return (intNumber%10).toString() + reverseInt(parseInt(intNumber/10))
}

toufaluanle · 2019-07-05T02:29:47Z

`var a = 1234;

function reverse(a){
var a = a.toString();
if(a.length==1){
return a
}else{
a = a.slice(-1)+reverse(a.slice(0,-1));
}
return a
}

reverse(a)`

ch8839 · 2019-07-05T03:13:32Z

思路：先将传入的参数转为string,取该字符串的第二位至末尾作为下一次递归函数的参数，判断当前字符串长度是否大于1，大于1则递归调用，等于1就返回这个字符串

function myReverse(num){ 
  let str = num.toString()
  let num2 = parseInt(str.slice(1))
  
  if(str.length >1){
    return myReverse(num2) + str[0]
  }else {
    return str
  } 
}

console.log(myReverse(1234))

lhyt · 2019-07-05T03:17:50Z

function reverse(num) {
  const str = `${num}`
  return str ? `${str % 10}${reverse(str.slice(0, str.length - 1))}` : ''
}

yaonote · 2019-07-05T04:33:25Z

cjfff · 2019-07-05T05:54:12Z

function renum(num) {
  num = num.toString()
  let rest = num.slice(0, -1),
    last = num.slice(-1);
  return last.length ? last + renum(rest) : ''
}

console.log(renum(12345678)); // 87654321

bran-nie · 2019-07-05T06:52:33Z

如果能传两个参数的话，这个算法将会更好。
尾递归调用。性能上好不少。

function func(v, r = '') {
    v += ''
    if (v.length < 1) return r
    r += v.slice(v.length-1)
    v = v.slice(0, v.length-1)
    return func(v, r)

}
func(1234)
// "4321"

YouziXR · 2019-07-05T07:27:18Z

如果能传两个参数的话，这个算法将会更好。
尾递归调用。性能上好不少。

function func(v, r = '') {
    v += ''
    if (v.length < 1) return r
    r += v.slice(v.length-1)
    v = v.slice(0, v.length-1)
    return func(v, r)

}
func(1234)
// "4321"

科里化一下就好了，在外面封装一层函数只接收参数v

xuxchao · 2019-07-06T01:37:17Z

function fun(num){
    let num1 = num / 10;
    let num2 = num % 10;
    if(num1<1){
        return num;
    }else{
        num1 = Math.floor(num1)
        return `${num2}${fun(num1)}`
    }
}
var a = fun(12345)
console.log(a)
console.log(typeof a)

输入的数字小于 10 的时候返回的是整形

momodiy · 2019-07-06T08:46:14Z

const convert = num =>
   (num / 10) | 0
     ? String(num % 10) + convert((num / 10) | 0 )
     : String(num)

let res = convert(12345)
console.log(res) // 54321
console.log(typeof res)// string

cycle07 · 2019-07-07T06:06:17Z

const calc = nums => nums.toString().split('').reduce((sum, num) => num + sum);

pengcc · 2019-07-07T07:47:53Z

贴上两种方法

// es6
const reverse1 = (num) => [...num.toString()].reverse().join('');

// no variable
var intReverseRecursive = function intReverseRecursive(num) {
    if (typeof num === 'number') {
        // add an anchor at the end
        return intReverseRecursive((num + '#').split(''));
    } else {
        // stop recursive
        if (num.indexOf('#') === 0) {
            num.shift();
            return num.join('');
        } else {
            num.splice(num.indexOf('#'), 0, num.shift());
            return intReverseRecursive(num);
        }
    }
};

zeybar · 2019-07-08T06:28:34Z

function reverseNumber (num) {
if ((num / 10 >> 0) === 0) {
return (‘’ + num)
}
return ('' + num % 10 + reverseNumber(num / 10 >> 0) )
}
或者简化成三行代码
function reverseNumber (num) {
return (num / 10 >> 0) === 0 ? ('' + num) : ('' + num % 10 + reverseNumber(num / 10 >> 0) )
}

你这个不行

5SSS · 2019-07-08T08:23:55Z

function intToString (int) {
	if (int >= 10) {
		return String(int % 10) + intToString(Math.floor(int / 10))
	}
	return String(int)
}

Zousdie · 2019-07-08T09:07:52Z

function doo (num) {
    const str = `${num}`;
    const temp = str.length > 2 ? doo(str.slice(1)) : str.slice(1);

    return `${temp}${str[0]}`
}

先转成字符串，然后根据长度递归调用

86XIng · 2019-07-08T13:25:26Z

const reverse = num=>
    num!=0 ? `${num % 10}${reverse(num / 10 >> 0)}` : ''

参考了各位大佬的思路，用%10来取最后一位，用/10来减少位数，用>>0来去除小数点，最后用es6的模板字符串来返回字符串类型的结果

momodiy · 2019-07-08T13:42:39Z

用 JavaScript 写一个函数，输入 int 型，返回整数逆序后的字符串。如：输入整型 1234，返回字符串“4321”。要求必须使用递归函数调用，不能用全局变量，输入函数必须只有一个参数传入，必须返回字符串。

int型整数也包含负数，发现大多数解法（包括我的）都只考虑传入数值为正数的情况，题干中也没有相关示例，求解 @yygmind

RanTing1992 · 2019-07-08T15:37:25Z

function fn1(num){
  var str = num.toString();
   let i = 0,newStr="";
   while(i<str.length){
      newStr+=str.slice(str.length-(i+1),str.length-i);
     i++
   }
return newStr;
}
fn1(1234)

FounderIsShadowWalker · 2019-07-09T03:12:17Z

var str = '12345';
function revert(str){
return str.length > 0 ? revert(str.slice(1)) + str[0] : str;
}

feikerwu · 2019-07-09T04:23:27Z

function fun(num){
    let num1 = num / 10;
    let num2 = num % 10;
    if(num1<1){
        return num;
    }else{
        num1 = Math.floor(num1)
        return `${num2}${fun(num1)}`
    }
}
var a = fun(12345)
console.log(a)
console.log(typeof a)

我觉着这个还是得加一些异常case，比如1000，或者负数

/**
 * @case 1234 => '4321'
 * @case 1000 => '1'
 * @case -1000 => '-1'
 * @case -1234 => '-4321'
 *
 * @param {number} number 传入的数值
 */
function solution(number) {
  if (number < 0) {
    return `-${solution(Math.abs(number))}`;
  }

  if (number < 10) {
    return `${number}`;
  }

  return `${number % 10 || ''}${solution(~~(number / 10))}`;
}

const assert = require('assert').strict;

assert.strictEqual(solution(1234), '4321');
assert.strictEqual(solution(1000), '1');
assert.strictEqual(solution(-1234), '-4321');
assert.strictEqual(solution(-1000), '-1');

Weathers0086 · 2019-07-09T06:13:23Z

function reverseNum(x){
if(x<0){
return '-'+reverseNum(Math.abs(x))
}else if(typeof x === 'number'){
return reverseNum(String(x))
}
return x.length===0?x:reverseNum(x.slice(1)) + x[0]
}
reverseNum(1234560)
reverseNum(-1234560)

FEJiuL · 2019-07-09T07:15:54Z

	function main(num){
		let [n, str] = ('' + num / 10).split(".");
                 return n > 0 ? str + main(n) : str;
	}

OhIAmFine · 2021-04-22T07:54:02Z

function reserveStr(num) {
  const str = num.toString(), len = str.length;
  if (len === 1 || !len) return str;
  if (len === 2) return str[1] + str[0];
  let res = "";
  for (let i = 0; i < len; i += 2) {
    res = reserveStr(str.slice(i, i + 2)) + res;
  }
  return res;
}

peterczg · 2021-05-04T06:48:10Z

function reverse(num){ return (num + "").length ? (reverse((num + "").slice(1)) + (num + "")[0]) : "" }

MrLeihe · 2021-05-11T08:36:28Z

function reverse(num) {
    // 得到个位数
    var mod = num % 10
    // 剩余位数
    var n = Math.floor(num / 10)
    if(n <= 0) {
        return num
    } else {
        return '' + mod + reverse(n) 
    }
}

nxt-hj · 2021-07-18T13:21:56Z

function toreverseString(int){
   if(!int&&int!==0) return ''
   var arr=int.toString().split('');
   var str = arr.pop()
   return str+toreverseString(arr.join(''))
}

635864859 · 2021-07-22T05:45:13Z

function fun(num){
    let num1 = num / 10;
    let num2 = num % 10;
    if(num1<1){
        return num;
    }else{
        num1 = Math.floor(num1)
        return `${num2}${fun(num1)}`
    }
}
var a = fun(12345)
console.log(a)
console.log(typeof a)

输入的数字小于 10 的时候返回的是整形

小于10的时候toString()一下么

printfshen · 2021-08-10T10:17:08Z

function core(input) {
  input = (input + "")
  let length = input.length
  let count = length - 1
  let result = ""
  return function f1() {
    result += input[count]
    if (count <= 0) {
      count = length - 1
      return result
    }
    count--
    return f1.call(null)
  }
}

core(123456789)()  // 987654321

iT-lijun · 2021-09-15T00:58:14Z

这可能是不依赖任何API的最佳实践了吧
const reverseInt = num => (num/10 >> 0) ? `${num%10}${reverseInt(num/10 >> 0)}` : `${num}`

lmfcz · 2021-09-23T01:52:04Z

// 用递归。第一步处理递归到底的条件第二步构建一个更小问题的解。第三步用更小问题的解构建出原问题的解
第一步很明显
第二步 reverse这个函数的宏观语意就是题目的意思，输入一个整数返回这个整数的逆序字符串。那么将原num % 10 后得到一个子整数即新的num reverse(num)就很明确的求出了原数字除了个位的逆序数
第三步将个位数与 reverse(num) 加起来构建原问题的解
const reverse = (num: number): string => {
if (num < 10) {
return num.toString()
}
const cur = (num % 10).toString()
num = Math.floor(num / 10)
return cur + reverse(num)
}

ganbowengo · 2021-10-13T12:53:31Z

function numberToString(num) {
let a = num % 10
let b = Math.floor(num/10)
if(b === 0) return a
return a + '' + numberToString(b)
}

xiangfei1 · 2021-10-26T06:42:59Z

function numToString(num, res = []) {
	if (num == '') return;
		let temp = Number(num) % 10;
		res.push(temp);
		let str = num.toString();
		if (res.length !== 0) {
			numToString(str.slice(0, -1), res)
		}
	return res.join('');
}

zhangtaolei · 2021-11-23T05:44:08Z

function reverse(num){
   if(num < 10) return num
   let str = ''
   str = `${num%10}${reverse(Math.floor(num/10))}`
  return str
 }

toSeeWorld · 2021-12-09T08:20:10Z

function reverse(val) {
if (!val) return ''
let str = val + ''
return reverse(+str.slice(1)) + str.charAt(0)
}

dymic-web · 2022-02-12T10:54:54Z

function reverse(num){
    var str=String(num)
    if(str.length===1){
        return str
    }

    return  reverse(str.slice(-1))+reverse(str.slice(0,-1))

}

Four-Names · 2022-02-26T03:53:17Z

function fun(int, x = int.toString().split(""), y = "") {
  if (x.length == 0) return y;
  else y = fun(undefined, x, x.shift()) + y;
  return y;
}
console.time("my");
console.log(fun(123212312312312313212312123321231231231));
console.timeEnd("my");

console.time("sys");
console.log(
  (123212312312312313212312123321231231231)
    .toString()
    .split("")
    .reverse()
    .join("")
);
console.timeEnd("sys");

liushan-github · 2022-03-08T01:55:56Z

function getRevString(num){
let numString= typeof num == 'string' ? num : num.toString();
if(numString<=1){
return numString
}else{
return ${numString.slice(numString.length- 1)}${getRevString(numString.slice(0,numString.length-1))}
}

}
console.log(getRevString(1234))

DearICE · 2022-03-14T03:41:56Z

function reverseN2S(num) {
  let str = num + '';
  let index = str.length - 1;
  return str[index] + (index > 0 ? reverseN2S(str.substring(0, index)) : '');
}

console.log(reverseN2S(1234));    // '4321'
console.log(reverseN2S(-1000));   // '0001-'

karl-barkmann · 2022-03-24T03:32:01Z

function input(num){
    const str = num.toString();
    function reverser(pos){
        if(pos<0){
            return '';
        }
        return str.charAt(pos)+reverser(--pos);
    }
    return reverser(str.length-1);
}

Twlig · 2022-04-13T13:36:49Z

function reverse(num) {
    if(num == undefined) {
        return
    }
    var str = num + ''
    if(str.length == 0) {
        return str
    }
    var s = str.charAt(str.length - 1)
    return s + reverse(str.substr(0, str.length - 1))
}

Margaux7 · 2022-04-17T14:00:50Z

function reverse(num) {
    num = typeof num === 'number' ? '' + num : num;
    if (!num.length) return '';
    
    return num.slice(-1) + reverse(num.slice(0, -1));
}

johnsoncheg · 2022-04-29T03:42:01Z

function input(num) {
	const strNum = String(num)
    
	if (strNum.length === 1) return strNum

	const d = Math.pow(10, strNum.length - 1)
	const res = Math.floor(num / d)
	const left = num % d
    
    console.log('res', res)
    console.log('d', d)
	return input(left) + res + ''
}

console.log(input(1234)) // '4321'

ShopkeeperEgg · 2022-06-03T10:28:06Z

function solve(int) {
  if (int === 0) return '';
  const tar = int % 10;
  // 如果可以不可以用Math
  // return tar.toString() + solve(int / 10 - int / 10 % 1);
  // 如果可以用parseInt
  return tar.toString() + solve(parseInt(int / 10));
}

FrankyYang · 2022-08-05T08:00:53Z

代码：const print = (num) => num < 10 ? '' + num : num % 10 + print(parseInt(num / 10))
测试：print(1234)

XW666 · 2022-08-22T01:55:29Z

//方式一
function numberReverse(num) {
    let m = num.toString().split(',')
    let res = []
    //递归
    let fun = function (list) {
      if (list.length > 0) {
        res.push(list.pop())
        fun(list)
      }
    }
    fun(m)
    return Number(res.join(''))
  }
//方式二
 var reverse = function (x) {
    let rev = 0;
    //方式二：递归的形式
    function fun(num) {
      if (num !== 0) {
        let num2 = num % 10;
        rev = rev * 10 + num2;
        //~~：去掉小数部分
        let num1 = ~~(num / 10);
        fun(num1)
      }
    }
    fun(x)
    return rev
  };

xuhen · 2022-11-30T09:26:16Z

function convert(n) {
    const str = n.toString();
    const helper = (s, i) => {
        if (i === s.length) return '';
        return helper(s, i + 1) + s[i];
    }
    return helper(str, 0);
}

ElderlyBoy · 2023-01-14T06:17:21Z

function fun(num){
    let num1 = num / 10;
    let num2 = num % 10;
    if(num1<1){
        return num;
    }else{
        num1 = Math.floor(num1)
        return `${num2}${fun(num1)}`
    }
}
var a = fun(12345)
console.log(a)
console.log(typeof a)

fun(1)// 1 number

benzhemin · 2023-03-12T13:27:26Z

const reverseToString = (value, res = []) => {
    if (value > 0) {
        const digital = value % 10;
        res.push(digital);
        return reverseToString(Math.floor(value / 10), res);
    }
    return res.join('');
}

const res = reverseToString(1234);

console.log(res);

qifengla · 2023-06-02T15:32:22Z

function reverseInteger(num) {
  if (num < 10) {
    return num.toString();
  }
  
  const lastDigit = num % 10;
  const remainingDigits = Math.floor(num / 10);
  
  return lastDigit.toString() + reverseInteger(remainingDigits);
}

Dylan0916 · 2023-08-25T15:09:19Z

function func(num) {
  const str = num + '';

  if (str.length === 1) {
    return str;
  }

  const newStr = str.slice(-1);
  const remainingStr = str.slice(0, -1);

  return newStr + func(remainingStr);
}

shawon1220 · 2023-12-20T10:28:34Z

(1234).toString().split('').reverse().join('')

Feng-K3 · 2024-04-09T08:36:53Z

其实就是一个整数反转啦

const reverse = (num) => {
    let res = 0;
    while (num) {
      res = res * 10 + (num % 10)
      num = ~~(num / 10) // 取整数
    }
    return res + ''; // 最后强转 String 类型
};
reverse(12345)  // '54321'

ghost2113 mentioned this issue Jul 8, 2019

``` 用到递归 #157

Closed

第 99 题：编程算法题 #153

第 99 题：编程算法题 #153

Comments

yygmind commented Jul 4, 2019

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