输入
array = [ 1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20 ]
输出将是
[ [ 1, 1, 1, 1 ], [ 2, 2, 2 ], 4, 5, 10, [ 20, 20 ], 391, 392, 591 ]
尝试了以下功能来创建结果：
var array = [ 1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
function answer(ArrayFromAbove) {
var length = array.length;
for (var i = 0; i < length; i++) {
for (var j = 0; j < (length - i - 1); j++) {
if (array[j] > array[j + 1]) {
var tmp = array[j];
array[j] = array[j + 1];
array[j + 1] = tmp;
}
}
}
}
answer(array);
console.log(array);
应返回：
[[1，1，1，1]，[2，2，2]，4，5，10，[20，20]，391，392，591]
答案 0 :(得分：2)
您可以考虑改用reduce，计算每个数字的出现次数，然后遍历已排序的条目并将值推入结果数组（作为数组，如果有多个值，或者只是一个纯数字（如果只有一个））：
const input = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20];
/* create an object like:
{
"1": 4,
"2": 3,
"4": 1,
"5": 1,
"10": 1,
"20": 2,
"391": 1,
"392": 1,
"591": 1
} */
const inputCounts = input.reduce((a, num) => {
a[num] = (a[num] || 0) + 1;
return a;
}, {});
const output = Object.entries(inputCounts)
// turn (string) key to number:
.map(([key, val]) => [Number(key), val])
.sort((a, b) => a[0] - b[0])
.reduce((a, [num, count]) => {
a.push(
count === 1
? num
: new Array(count).fill(num)
);
return a;
}, []);
console.log(output);
答案 1 :(得分：1)
首先使用Set和 Array.prototype.sort() 删除重复项。获取数组中的数字计数，并fill()和新的Array和push()进行输出。
const array=[1,2,4,591,392,391,2,5,10,2,1,1,1,20,20]
function count(arr,num){
return Array(arr.filter(x => x === num).length).fill(num);
}
const unique = [...new Set(array)].sort((a,b) => a -b);
const output = unique.map(un => {
let arr = count(array,un);
return (arr.length > 1) ? arr : un;
})
console.log(output)
upper方法正在遍历数组。您可以按照以下方法循环两次来完成
let array=[1,2,4,591,392,391,2,5,10,2,1,1,1,20,20]
array = array.sort((a,b) => a-b);
const result = [];
let temp = [];
for(let i = 0;i<array.length + 1;++i){
if(array[i - 1] === array[i] || i === 0){
temp.push(array[i]);
}
else{
result.push((temp.length === 1) ? temp[0] : temp);
temp = [array[i]];
}
}
//result.push(temp)
console.log(result);
答案 2 :(得分：0)
您可以计算发生的次数，然后根据计数创建数组：
const counts = new Map;
for(const value of array.sort())
counts.set(value, (counts.get(value) || 0) + 1);
const result = [];
for(const [count, value] of counts.entries())
result.push(new Array(count).fill(value));
答案 3 :(得分：0)
您可以尝试关注
let array = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
let result = array.sort((a,b) => a-b).reduce((a,c) => {
if(a.length) {
let lastVal = a[a.length-1];
if(Array.isArray(lastVal)) {
if(lastVal[0] == c) lastVal.push(c);
else a.push(c);
} else if(lastVal == c) a[a.length-1] = [c,c];
else a.push(c);
}
else a.push(c);
return a;
}, []);
console.log(result);
答案 4 :(得分：0)
您可以根据数字的出现次数使用减少和填充数组
var array = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
let reduced = array.reduce((op,inp)=>((op[inp] ? op[inp]++ : (op[inp]=1)),op),{})
let op = Object.keys(reduced).map(e=>(reduced[e] === 1 ? e : new Array(reduced[e]).fill(e)))
.sort((a,b)=> (Array.isArray(a) ? a[0] : a) - (Array.isArray(b) ? b[0] : b) )
console.log(op)
答案 5 :(得分：0)
对数组进行排序，将所有数字分组为一个子数组，然后仅用该元素中的一个元素替换这些数组。
var array = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
function answer(arr){
arr.sort((a,b) => a - b);
var tempResult = [];
while(arr.length > 0){
var val = arr.shift();
if(tempResult.length == 0 || tempResult[tempResult.length-1][0] !== val){
tempResult.push([val]);
}else{
tempResult[tempResult.length-1].push(val);
}
}
while(tempResult.length > 0){
var subArr = tempResult.shift();
if(subArr.length == 1) arr.push(subArr[0]);
else arr.push(subArr);
}
}
answer(array);
console.log(JSON.stringify(array));
答案 6 :(得分：0)
使用<?php
namespace Tuurbo\Spreedly;
use GuzzleHttp\Client as Guzzle;
class Spreedly
{
protected $config;
public function __construct($config)
{
$this->setConfig($config);
}
public function gateway($token = null)
{
return new Gateway($this->client(), $this->config, $token);
}
public function payment($paymentToken = null)
{
return new Payment($this->client(), $this->config, $paymentToken, $this->gateway()->getToken());
}
public function transaction($token = null)
{
return new Transaction($this->client(), $this->config, $token);
}
public function setConfig(array $config)
{
$this->config = $config;
$this->validateConfig(); // this is where I have problem
return $this;
}
protected function validateConfig()
{
// if (!isset($this->config['key'])) {
if(!isset(($this->config['spreedly']['key']))){ // this is suppose to be fail
throw new Exceptions\InvalidConfigException(); // this is thrown
}
if (!isset($this->config['secret'])) {
throw new Exceptions\InvalidConfigException();
}
}
public function client()
{
return new Client(new Guzzle(), $this->config);
}
}
和Array.reduce()将重复项累积到一个对象中，然后根据重复项的键将元素过滤并映射到一个新数组中。
Array.sort()
答案 7 :(得分：0)
这是我的实现方式
var array = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
function answer (input) {
return input.sort((a,b) => a - b)
.reduce((collect, next) => {
// First item, just add it
if (!collect.length) {
collect.push(next)
return collect
}
let prev = collect.pop()
if (Array.isArray(prev)) {
let [ num ] = prev
if (num === next) {
prev.push(next)
collect.push(prev)
return collect
} else {
collect.push(prev, next)
return collect
}
} else if ( prev === next ) {
collect.push([prev, next])
return collect
} else {
collect.push(prev, next)
return collect
}
}, []);
}
console.log(answer(array));
答案 8 :(得分：0)
var array = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
var resultArray = []
var subArray = []
var length = array.length;
array.sort(function sortArray(a,b) {
return a - b;
});
for(var i = 0; i < length; i++)
{
if(subArray.length == 0 || subArray[0] == array[i])
{
subArray.push(array[i]);
}
else if(subArray.length > 1 && array[i] != subArray[0])
{
resultArray.push(subArray);
subArray = [array[i]];
}
else if(subArray.length == 1 && array[i] != subArray[0])
{
resultArray.push(subArray[0]);
subArray = [array[i]];
}
}
if(subArray.length == 1)
{
resultArray.push(subArray[0]);
}
else
{
resultArray.push(subArray);
}
console.log(resultArray);
//return(resultArray);