如何在Javascript中“正确”相交两个数组[关闭]

Question

有没有合适的方法在 Javascript 中相交两个数组？

我试图以正确的方式使两个数组相交，但我发现了一些困难

我的输入未按此处假设的方式排序 Simplest code for array intersection in javascript

我的代码：

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function(nums1, nums2) {
    var t, a = nums1,b = nums2;
    if (b.length > a.length) t = b, b = a, a = t;
    return a.filter(x => b.includes(x))
};
console.log(intersect([1, 2], [1, 1])); //[1] correct
console.log(intersect([1, 1], [1, 2])); //[1,1] wrong
console.log(intersect([1], [1, 1])); //[1,1] wrong
console.log(intersect([1, 1, 1], [1, 1])); //[1,1,1] wrong

并从该答案中尝试了该算法： Finding the intersection of two arrays in Javascript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function(nums1, nums2) {
  var A=nums1,B=nums2;
  var m = A.reduce(function(m, v) { m[v] = 1; return m; }, {});
  return B.filter(function(v) { return m[v]; });
};
console.log(intersect([1, 2], [1, 1])); //[1,1] wrong
console.log(intersect([1, 1], [1, 2])); //[1] right
console.log(intersect([1], [1, 1])); //[1,1] wrong
console.log(intersect([1, 1, 1], [1, 1])); //[1,1] right

也试过这个：

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function(nums1, nums2) {
  return nums1.filter((n) => nums2.indexOf(n) !== -1);
};
console.log(intersect([1, 2], [1, 1])); //[1] right
console.log(intersect([1, 1], [1, 2])); //[1,1] wrong
console.log(intersect([1], [1, 1])); //[1] right
console.log(intersect([1, 1, 1], [1, 1])); //[1,1,1] wrong

还：

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function(nums1, nums2) {
  var a = nums1,b=nums2;
  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;
};
console.log(intersect([1, 2], [1, 1])); //[1] right
console.log(intersect([1, 1], [1, 2])); //[1] right
console.log(intersect([1], [1, 1])); //[1] right
console.log(intersect([1, 1, 1], [1, 1])); //[1] right
console.log(intersect([2,1],[1,1])); //[] wrong

我错在哪里了？

Answer 1

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function(nums1, nums2) {
    var arr = [], ind;
    while (nums1.length) {
        ind = nums2.indexOf(nums1.shift());
        if (ind > -1) {
            arr.push(nums2.splice(ind, 1)[0]);
        }
    }
    return arr;
};

Answer 2

您可以使用Map 并减少计数（如果找到用于过滤）。

var intersect = function(nums1, nums2) {
    var m = new Map();
    
    nums1.forEach(a => m.set(a, (m.get(a) || 0) + 1));
    return nums2.filter(a => m.get(a) && m.set(a, m.get(a) - 1));
};

console.log(intersect([1, 2], [1, 1]));    // [1]
console.log(intersect([1, 1], [1, 2]));    // [1]
console.log(intersect([1], [1, 1]));       // [1]
console.log(intersect([1, 1, 1], [1, 1])); // [1, 1]

Answer 3

另一个实现，使用您的方法 1，但使用 ES 6 Set 删除重复项

    /**
     * @param {number[]} nums1
     * @param {number[]} nums2
     * @return {number[]}
     */
    var intersect = function(nums1, nums2) {
         var result = nums1.filter(x => nums2.includes(x));
         return [...new Set(result)];
    };

    console.log(intersect([1, 2], [1, 1])); //[1] correct
    console.log(intersect([1, 1], [1, 2])); //[1] correct
    console.log(intersect([1], [1, 1])); //[1] correct
    console.log(intersect([1, 1, 1], [1, 1])); //[1] correct
    console.log(intersect([1, 45, 143, 76, 11], [761,76, 11, 1])); //[1,76,11] correct