TA貢獻(xiàn)1821條經(jīng)驗(yàn) 獲得超5個(gè)贊
使用的組合Array.prototype.filter和Array.prototype.indexOf：
array1.filter(value => -1 !== array2.indexOf(value))
或者正如vrugtehagel在評(píng)論中所建議的那樣，你可以使用更新的更Array.prototype.includes簡(jiǎn)單的代碼：
array1.filter(value => array2.includes(value))
對(duì)于舊瀏覽器：
array1.filter(function(n) {
return array2.indexOf(n) !== -1;});
TA貢獻(xiàn)1850條經(jīng)驗(yàn) 獲得超11個(gè)贊
破壞性似乎最簡(jiǎn)單，特別是如果我們可以假設(shè)輸入已排序：
/* 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;}非破壞性必須是一個(gè)更復(fù)雜的頭發(fā)，因?yàn)槲覀儽仨毟櫵饕?/p>
/* 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;}
TA貢獻(xiàn)1875條經(jīng)驗(yàn) 獲得超3個(gè)贊
如果您的環(huán)境支持ECMAScript 6 Set，那么一種簡(jiǎn)單且有效的（參見規(guī)范鏈接）方式：
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，但可讀性更低（也沒有創(chuàng)建額外的交集Set）：
function intersect(a, b) {
return [...new Set(a)].filter(x => new Set(b).has(x));}避免新Set的b每次：
function intersect(a, b) {
var setB = new Set(b);
return [...new Set(a)].filter(x => setB.has(x));}請(qǐng)注意，使用集合時(shí)，您將只獲得不同的值，因此new Set[1,2,3,3].size計(jì)算結(jié)果為3。
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