使用的组合Array.prototype.filter和Array.prototype.includes：

const filteredArray = array1.filter(value => array2.includes(value));

对于旧浏览器，带Array.prototype.indexOf和不带箭头功能：

var filteredArray = array1.filter(function(n) {
    return array2.indexOf(n) !== -1;
});

注意！二者.includes并.indexOf在内部通过使用阵列中的元素进行比较===，因此，如果该数组包含对象它将只比较对象引用（不是他们的内容）。如果要指定自己的比较逻辑，请Array.prototype.some改用。

破坏性似乎最简单，特别是如果我们可以假设输入已排序：

/* destructively finds the intersection of 
 * two arrays in a simple fashion.  
 *
 * PARAMS
 *  a - first array, must already be sorted
 *  b - second array, must already be sorted
 *
 * NOTES
 *  State of input arrays is undefined when
 *  the function returns.  They should be 
 *  (prolly) be dumped.
 *
 *  Should have O(n) operations, where n is 
 *    n = MIN(a.length, b.length)
 */
function intersection_destructive(a, b)
{
  var result = [];
  while( a.length > 0 && b.length > 0 )
  {  
     if      (a[0] < b[0] ){ a.shift(); }
     else if (a[0] > b[0] ){ b.shift(); }
     else /* they're equal */
     {
       result.push(a.shift());
       b.shift();
     }
  }

  return result;
}

非破坏性必须更复杂，因为我们必须跟踪索引：

/* finds the intersection of 
 * two arrays in a simple fashion.  
 *
 * PARAMS
 *  a - first array, must already be sorted
 *  b - second array, must already be sorted
 *
 * NOTES
 *
 *  Should have O(n) operations, where n is 
 *    n = MIN(a.length(), b.length())
 */
function intersect_safe(a, b)
{
  var ai=0, bi=0;
  var result = [];

  while( ai < a.length && bi < b.length )
  {
     if      (a[ai] < b[bi] ){ ai++; }
     else if (a[ai] > b[bi] ){ bi++; }
     else /* they're equal */
     {
       result.push(a[ai]);
       ai++;
       bi++;
     }
  }

  return result;
}

如果您的环境支持ECMAScript 6 Set，一种简单且据称有效（参见规范链接）的方法：

function intersect(a, b) {
  var setA = new Set(a);
  var setB = new Set(b);
  var intersection = new Set([...setA].filter(x => setB.has(x)));
  return Array.from(intersection);
}

更短，但可读性较差（也没有创建额外的交叉点Set）：

function intersect(a, b) {
  var setB = new Set(b);
  return [...new Set(a)].filter(x => setB.has(x));
}

请注意，使用集合时，您只会得到不同的值，因此new Set([1, 2, 3, 3]).size计算结果为3。

使用 Underscore.js或lodash.js

_.intersection( [0,345,324] , [1,0,324] )  // gives [0,324]

// Return elements of array a that are also in b in linear time:
function intersect(a, b) {
  return a.filter(Set.prototype.has, new Set(b));
}

// Example:
console.log(intersect([1,2,3], [2,3,4,5]));

我推荐上述简洁的解决方案，它在大输入上优于其他实现。如果小输入的性能很重要，请检查下面的替代方案。

替代方案和性能比较：

有关替代实现，请参阅以下代码段，并检查https://jsperf.com/array-intersection-comparison以进行性能比较。

function intersect_for(a, b) {
  const result = [];
  const alen = a.length;
  const blen = b.length;
  for (let i = 0; i < alen; ++i) {
    const ai = a[i];
    for (let j = 0; j < blen; ++j) {
      if (ai === b[j]) {
        result.push(ai);
        break;
      }
    }
  } 
  return result;
}

function intersect_filter_indexOf(a, b) {
  return a.filter(el => b.indexOf(el) !== -1);
}

function intersect_filter_in(a, b) {
  const map = b.reduce((map, el) => {map[el] = true; return map}, {});
  return a.filter(el => el in map);
}

function intersect_for_in(a, b) {
  const result = [];
  const map = {};
  for (let i = 0, length = b.length; i < length; ++i) {
    map[b[i]] = true;
  }
  for (let i = 0, length = a.length; i < length; ++i) {
    if (a[i] in map) result.push(a[i]);
  }
  return result;
}

function intersect_filter_includes(a, b) {
  return a.filter(el => b.includes(el));
}

function intersect_filter_has_this(a, b) {
  return a.filter(Set.prototype.has, new Set(b));
}

function intersect_filter_has_arrow(a, b) {
  const set = new Set(b);
  return a.filter(el => set.has(el));
}

function intersect_for_has(a, b) {
  const result = [];
  const set = new Set(b);
  for (let i = 0, length = a.length; i < length; ++i) {
    if (set.has(a[i])) result.push(a[i]);
  }
  return result;
}

在 Firefox 53 中的结果：

• 大型阵列（10,000 个元素）上的操作数/秒：

  filter + has (this)               523 (this answer)
  for + has                         482
  for-loop + in                     279
  filter + in                       242
  for-loops                          24
  filter + includes                  14
  filter + indexOf                   10
  

• 小阵列（100 个元素）上的操作数/秒：

  for-loop + in                 384,426
  filter + in                   192,066
  for-loops                     159,137
  filter + includes             104,068
  filter + indexOf               71,598
  filter + has (this)            43,531 (this answer)
  filter + has (arrow function)  35,588