另一种解决问题的方法

function diffArray(arr1, arr2) {
    return arr1.concat(arr2).filter(function (val) {
        if (!(arr1.includes(val) && arr2.includes(val)))
            return val;
    });
}

diffArray([1, 2, 3, 7], [3, 2, 1, 4, 5]);    // return [7, 4, 5]

同样，你可以使用箭头函数语法:

const diffArray = (arr1, arr2) => arr1.concat(arr2)
    .filter(val => !(arr1.includes(val) && arr2.includes(val)));

diffArray([1, 2, 3, 7], [3, 2, 1, 4, 5]);    // return [7, 4, 5]

基于思考者的答案，但允许重复。

映射在映射值出现时增加它们，如果它们在另一个数组中，则减少它们。

任何剩余的部分都将包括在差额中。

function diff(curr, prev) { let a = curr.split('').sort(), b = prev.split('').sort(), c = arrDiff(a, b); console.log(JSON.stringify(a), '-', JSON.stringify(b), '=', JSON.stringify(c)); return c; } function arrDiff(larger, smaller) { var entries = {}; for (var i = 0; i < larger.length; i++) { entries[larger[i]] = (entries[larger[i]] || 0) + 1; } for (var i = 0; i < smaller.length; i++) { if (entries[smaller[i]]) { entries[smaller[i]] -= 1; } else { entries[smaller[i]] = (entries[smaller[i]] || 0) + 1; } } return Object.keys(entries).sort().reduce((diff, key) => { if (entries[key] > 0) { for (var i = 0; i < entries[key]; i++) { diff.push(key); } } return diff; }, []); } // Smaller is a subset of Larger console.log('Result:', JSON.stringify(diff('ENLIGHTEN', 'LENGTHEN'))); // [ I ] console.log('Result:', JSON.stringify(diff('STRENGTH', 'TENTHS'))); // [ G, R ] // Both have a unique value console.log('Result:', JSON.stringify(diff('BUBBLE', 'RUBBLE'))); // [ B, R ] .as-console-wrapper { top: 0; max-height: 100% !important; }

简单地比较所有值，并返回数组与不重复的值。

var main = [9, '$', 'x', 'r', 3, 'A', '#', 0, 1];

var arr0 = ['Z', 9, 'e', '$', 'r'];
var arr1 = ['x', 'r', 3, 'A', '#'];
var arr2 = ['m', '#', 'a', 0, 'r'];
var arr3 = ['$', 1, 'n', '!', 'A'];


Array.prototype.diff = function(arrays) {
    var items = [].concat.apply(this, arguments);
    var diff = [].slice.call(items), i, l, x, pos;

    // go through all items
    for (x = 0, i = 0, l = items.length; i < l; x = 0, i++) {
        // find all positions
        while ((pos = diff.indexOf(items[i])) > -1) {
            // remove item + increase found count
            diff.splice(pos, 1) && x++;
        }
        // if item was found just once, put it back
        if (x === 1) diff.push(items[i]);
    }
    // get all not duplicated items
    return diff;
};

main.diff(arr0, arr1, arr2, arr3).join(''); // returns "Zeman!"

[].diff(main, arr0, arr1, arr2, arr3).join(''); // returns "Zeman!"

只是想…为了一个挑战;-)这个工作…(对于字符串，数字等的基本数组)没有嵌套数组

function diffArrays(arr1, arr2, returnUnion){
  var ret = [];
  var test = {};
  var bigArray, smallArray, key;
  if(arr1.length >= arr2.length){
    bigArray = arr1;
    smallArray = arr2;
  } else {
    bigArray = arr2;
    smallArray = arr1;
  }
  for(var i=0;i<bigArray.length;i++){
    key = bigArray[i];
    test[key] = true;
  }
  if(!returnUnion){
    //diffing
    for(var i=0;i<smallArray.length;i++){
      key = smallArray[i];
      if(!test[key]){
        test[key] = null;
      }
    }
  } else {
    //union
    for(var i=0;i<smallArray.length;i++){
      key = smallArray[i];
      if(!test[key]){
        test[key] = true;
      }
    }
  }
  for(var i in test){
    ret.push(i);
  }
  return ret;
}

array1 = "test1", "test2","test3", "test4", "test7"
array2 = "test1", "test2","test3","test4", "test5", "test6"
diffArray = diffArrays(array1, array2);
//returns ["test5","test6","test7"]

diffArray = diffArrays(array1, array2, true);
//returns ["test1", "test2","test3","test4", "test5", "test6","test7"]

注意排序可能不会像上面提到的那样…但是如果需要的话，可以在数组上调用.sort()对其进行排序。

随着ES6的到来，带有sets和splat操作符(当时只在Firefox中工作，请检查兼容性表)，你可以编写以下一行代码:

var a = ['a', 'b', 'c', 'd'];
var b = ['a', 'b'];
var b1 = new Set(b);
var difference = [...new Set(a.filter(x => !b1.has(x)))];

结果是["c"， "d"]。

    function arrayDiff(a, b) {
      return a.concat(b).filter(val => !(b.includes(val)));
      //(or) return a.concat(b).filter(val => !(a.includes(val) && b.includes(val)));
    }