用javascript实现数组交叉的最简单、无库代码是什么?我想写

intersection([1,2,3], [2,3,4,5])

并获得

[2, 3]

当前回答

这是一个现代和简单的ES6方式来做,也非常灵活。 它允许您指定多个数组作为与主题数组进行比较的数组,并且可以在包含和独占模式下工作。

// =======================================
// The function
// =======================================

function assoc(subjectArray, otherArrays, { mustBeInAll = true } = {}) {
  return subjectArray.filter((subjectItem) => {
    if (mustBeInAll) {
      return otherArrays.every((otherArray) =>
        otherArray.includes(subjectItem)
      );
    } else {
      return otherArrays.some((otherArray) => otherArray.includes(subjectItem));
    }
  });
}

// =======================================
// The usage
// =======================================

const cheeseList = ["stilton", "edam", "cheddar", "brie"];
const foodListCollection = [
  ["cakes", "ham", "stilton"],
  ["juice", "wine", "brie", "bread", "stilton"]
];

// Output will be: ['stilton', 'brie']
const inclusive = assoc(cheeseList, foodListCollection, { mustBeInAll: false }),

// Output will be: ['stilton']
const exclusive = assoc(cheeseList, foodListCollection, { mustBeInAll: true })

实例:https://codesandbox.io/s/zealous-butterfly-h7dgf?fontsize=14&hidenavigation=1&theme=dark

其他回答

解决它 从索引0开始逐一检查,然后创建一个新数组。

像这样的东西,不过测试不太好。

function intersection(x,y){
 x.sort();y.sort();
 var i=j=0;ret=[];
 while(i<x.length && j<y.length){
  if(x[i]<y[j])i++;
  else if(y[j]<x[i])j++;
  else {
   ret.push(x[i]);
   i++,j++;
  }
 }
 return ret;
}

alert(intersection([1,2,3], [2,3,4,5]));

PS:该算法仅适用于数字和普通字符串,任意对象数组的交集可能无法工作。

在coffescript中N个数组的交集

getIntersection: (arrays) ->
    if not arrays.length
        return []
    a1 = arrays[0]
    for a2 in arrays.slice(1)
        a = (val for val in a1 when val in a2)
        a1 = a
    return a1.unique()

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]

破坏性似乎是最简单的,特别是如果我们可以假设输入是排序的:

/* 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;
}

非破坏性的要稍微复杂一点,因为我们要跟踪指标:

/* 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;
}

“filter”和“indexOf”在IE中的Array上不支持。这个怎么样:

var array1 = [1, 2, 3];
var array2 = [2, 3, 4, 5];

var intersection = [];
for (i in array1) {
    for (j in array2) {
        if (array1[i] == array2[j]) intersection.push(array1[i]);
    }
}