已经有很多答案了，但不幸的是，有些太长了，有些太短了，但对我来说太神秘了，而另一些则超出了我的知识范围……不过，我真的很喜欢我提出的这个解决方案。希望它仍然对一些人有帮助!

尽管最初的帖子说他/她实际上不需要重复的索引，也不需要重复多少次，但我认为仍然需要清楚地计算它们。

带有注释的代码。

function findDuplicates(array, count = {}) {
  // with count declared in the parameter, initialized as an empty object, 
  // it can store the counts of all elements in array  
  
  // using the forEach loop to iterate through the input array, 
  // also using the conditional ternary operators 
  // (works just like a normal if-else statement, but just a bit cleaner)
  // we can store all occurrences of each element from array in count
  array.forEach(el => count[el] ? count[el]++ : count[el] = 1)
  
  // using Object.keys, we get an array of all keys from count (all numbers) 
  // (sorted as well, though of no specific importance here)
  // using filter to find all elements with a count (value) > 1 (duplicates!)
  return Object.keys(count).filter(key => count[key] > 1);
}

只有代码(带有测试用例)。

函数findduplicate(数组，count = {}) { 数组中。forEach(el => count[el] ?Count [el]++: Count [el] = 1); 返回种(计数)。Filter (key => count[key] > 1); ｝ 让arr1 = [9,9,111, 2,3,4,4,5,7]; 让arr2 = [1,6,7,3,6,8,1,3,4,5,1,7,2,6]; console.log (findDuplicates (arr1));// => ['4'， '9'] console.log (findDuplicates (arr2));// => ['1'， '3'， '6'， '7']

ES6语法的简单代码(返回重复的排序数组):

let duplicates = a => {d=[]; a.sort((a,b) => a-b).reduce((a,b)=>{a==b&&!d.includes(a)&&d.push(a); return b}); return d};

使用方法:

duplicates([1,2,3,10,10,2,3,3,10]);

这是我的建议(ES6):

let a = [1, 2, 3, 4, 2, 2, 4, 1, 5, 6]
let b = [...new Set(a.sort().filter((o, i) => o !== undefined && a[i + 1] !== undefined && o === a[i + 1]))]

// b is now [1, 2, 4]

你可以使用排序、筛选和集合来做到这一点。

Var数= [1,2,3,4,5,6,7,8,1,2,3,4,5,3,4]; var numbersSorted = numbers.sort(); 令result = numbers。Filter ((e, i) => numbers[i] == numbers[i+1]); 结果=[…]新设置(结果)]; console.log(结果);

基于@ blumoon但更短，返回所有副本一次!

function checkDuplicateKeys(arr) {
    const counts = {}
    return arr.filter((item) => {
        counts[item] = counts[item] || 1
        if (counts[item]++ === 2) return true
    })
}

// [1,2,2,2,2,2,2] => [1,2]
// ['dog', 'dog', 'cat'] => ['dog']

魔法

a.filter(( t={}, e=>!(1-(t[e]=++t[e]|0)) )) 

O (n)的性能;我们假设你的数组在a中，它包含可以以唯一方式转换. tostring()的元素(这是由JS在t[e]中隐式完成的)，例如numbers=[4,5,4]， strings=["aa"，"bb"，"aa"]， arraysNum=[[1,2,3]，[43,2,3]，[1,2,3]]。这里有解释，这里有唯一值

var a1 = [[2, 17], [2, 17], [2, 17], [1, 12], [5, 9], [1, 12], [6,2], [1, 12]]; var a2 =[“迈克”,“亚当”,“马特”、“南希”,“亚当”,“珍妮”,“南希”,“卡尔”); Var a3 = [5,6,4,9,2,3,5,3,4,1,5,4,9]; 让nd = (a) = > a.filter ((t = {}, e = > ! (1 - (t [e] = + + t [e] | 0)))) / /打印 let c= x => console.log(JSON.stringify(x)); C (nd(a1)); C (nd(a2)); C (nd(a3));