//在线性时间内返回数组a中也在b中的元素: 函数相交(a, b) { 返回a.filter (Set.prototype。new Set(b)); ｝ / /例如: console.log(相交([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。如果在小输入上的性能很重要，请检查下面的替代方案。

备选方案和性能比较:

有关替代实现，请参阅下面的代码片段，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; }

Firefox 53的结果:

Ops/sec on large arrays (10,000 elements): 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 Ops/sec on small arrays (100 elements): 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

另一种可以同时处理任意数量数组的索引方法:

// Calculate intersection of multiple array or object values.
function intersect (arrList) {
    var arrLength = Object.keys(arrList).length;
        // (Also accepts regular objects as input)
    var index = {};
    for (var i in arrList) {
        for (var j in arrList[i]) {
            var v = arrList[i][j];
            if (index[v] === undefined) index[v] = 0;
            index[v]++;
        };
    };
    var retv = [];
    for (var i in index) {
        if (index[i] == arrLength) retv.push(i);
    };
    return retv;
};

它只适用于可以作为字符串计算的值，你应该将它们作为一个数组传递:

intersect ([arr1, arr2, arr3...]);

.．.但它透明地接受对象作为参数或任何要交叉的元素(总是返回公共值的数组)。例子:

intersect ({foo: [1, 2, 3, 4], bar: {a: 2, j:4}}); // [2, 4]
intersect ([{x: "hello", y: "world"}, ["hello", "user"]]); // ["hello"]

编辑:我只是注意到，这是，在某种程度上，有点bug。

也就是说:我在编码时认为输入数组本身不能包含重复(正如所提供的示例那样)。

但如果输入数组恰好包含重复，就会产生错误的结果。示例(使用下面的实现):

intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]);
// Expected: [ '1' ]
// Actual: [ '1', '3' ]

幸运的是，这很容易通过添加二级索引来解决。那就是:

变化:

        if (index[v] === undefined) index[v] = 0;
        index[v]++;

by:

        if (index[v] === undefined) index[v] = {};
        index[v][i] = true; // Mark as present in i input.

,:

         if (index[i] == arrLength) retv.push(i);

by:

         if (Object.keys(index[i]).length == arrLength) retv.push(i);

完整的例子:

// Calculate intersection of multiple array or object values.
function intersect (arrList) {
    var arrLength = Object.keys(arrList).length;
        // (Also accepts regular objects as input)
    var index = {};
    for (var i in arrList) {
        for (var j in arrList[i]) {
            var v = arrList[i][j];
            if (index[v] === undefined) index[v] = {};
            index[v][i] = true; // Mark as present in i input.
        };
    };
    var retv = [];
    for (var i in index) {
        if (Object.keys(index[i]).length == arrLength) retv.push(i);
    };
    return retv;
};

intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]); // [ '1' ]

这是我使用的一个非常简单的实现。它是无损的，也确保不复制整个。

Array.prototype.contains = function(elem) {
    return(this.indexOf(elem) > -1);
};

Array.prototype.intersect = function( array ) {
    // this is naive--could use some optimization
    var result = [];
    for ( var i = 0; i < this.length; i++ ) {
        if ( array.contains(this[i]) && !result.contains(this[i]) )
            result.push( this[i] );
    }
    return result;
}

ES2015的函数式方法

函数式方法必须考虑只使用没有副作用的纯函数，每个函数只与单个作业有关。

这些限制增强了所涉及函数的可组合性和可重用性。

//小的，可重用的辅助函数 const createSet = xs => new Set(xs); Const filter = f => xs => xs.filter(apply(f)); Const apply = f => x => f(x); / /十字路口 Const相交= xs => ys => { const zs =创建集(ys); 返回过滤器(x => zs.has(x)) ？ 真正的 :假 ) (x); }; //模拟数据 Const xs = [1,2,2,3,4,5]; Const ys = [0,1,2,3,3,3,6,7,8,9]; //运行 Console.log (intersect(xs) (ys));

请注意，使用本机Set类型，这有一个优点 查找性能。

避免重复

显然，第一个数组中重复出现的项将被保留，而第二个数组将被去重。这可能是也可能不是理想的行为。如果你需要一个唯一的结果，只需对第一个参数应用重复数据删除:

// auxiliary functions const apply = f => x => f(x); const comp = f => g => x => f(g(x)); const afrom = apply(Array.from); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // intersection const intersect = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? true : false ) (xs); }; // de-duplication const dedupe = comp(afrom) (createSet); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // unique result console.log( intersect(dedupe(xs)) (ys) );

计算任意数量数组的交集

如果你想计算任意数量的数组的交点，只需用compose intersect和foldl。这是一个方便函数:

// auxiliary functions const apply = f => x => f(x); const uncurry = f => (x, y) => f(x) (y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); const foldl = f => acc => xs => xs.reduce(uncurry(f), acc); // intersection const intersect = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? true : false ) (xs); }; // intersection of an arbitrarily number of Arrays const intersectn = (head, ...tail) => foldl(intersect) (head) (tail); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; const zs = [0,1,2,3,4,5,6]; // run console.log( intersectn(xs, ys, zs) );

我认为在内部使用一个对象可以帮助计算，也可以提高性能。

//方法维护每个元素的计数，也适用于负元素

function intersect(a,b){
    
    const A = {};
    a.forEach((v)=>{A[v] ? ++A[v] : A[v] = 1});
    const B = {};
    b.forEach((v)=>{B[v] ? ++B[v] : B[v] = 1});
    const C = {};
    Object.entries(A).map((x)=>C[x[0]] = Math.min(x[1],B[x[0]]))
    return Object.entries(C).map((x)=>Array(x[1]).fill(Number(x[0]))).flat();
}
const x = [1,1,-1,-1,0,0,2,2];
const y = [2,0,1,1,1,1,0,-1,-1,-1];
const result = intersect(x,y);
console.log(result);  // (7) [0, 0, 1, 1, 2, -1, -1]

通过对数据的一些限制，您可以在线性时间内完成!

对于正整数:使用一个数组将值映射到“已见/未见”布尔值。

function intersectIntegers(array1,array2) { 
   var seen=[],
       result=[];
   for (var i = 0; i < array1.length; i++) {
     seen[array1[i]] = true;
   }
   for (var i = 0; i < array2.length; i++) {
     if ( seen[array2[i]])
        result.push(array2[i]);
   }
   return result;
}

对于对象也有类似的技术:取一个虚拟键，为array1中的每个元素设置为“true”，然后在array2的元素中寻找这个键。完事后收拾一下。

function intersectObjects(array1,array2) { 
   var result=[];
   var key="tmpKey_intersect"
   for (var i = 0; i < array1.length; i++) {
     array1[i][key] = true;
   }
   for (var i = 0; i < array2.length; i++) {
     if (array2[i][key])
        result.push(array2[i]);
   }
   for (var i = 0; i < array1.length; i++) {
     delete array1[i][key];
   }
   return result;
}

当然，你需要确保这个键之前没有出现过，否则你会破坏你的数据…