使用额外的内存来做到这一点。这样你可以用更少的时间复杂度来求解，O(n)而不是O(n *n)

function getDiff(arr1,arr2){
let k = {};
let diff = []
arr1.map(i=>{
    if (!k.hasOwnProperty(i)) {
        k[i] = 1
    }
}
)
arr2.map(j=>{
    if (!k.hasOwnProperty(j)) {
        k[j] = 1;
    } else {
        k[j] = 2;
    }
}
)
for (var i in k) {
    if (k[i] === 1)
        diff.push(+i)
}
return diff
}
getDiff([4, 3, 52, 3, 5, 67, 9, 3],[4, 5, 6, 75, 3, 334, 5, 5, 6])

const difference = function (baseArray, arrayToCampare, callback = (a, b) => a!== b) {
  if (!(arrayToCampare instanceof Array)) {
    return baseArray;
  }
  return baseArray.filter(baseEl =>
    arrayToCampare.every(compareEl => callback(baseEl, compareEl)));
}

纯JavaScript解决方案(没有库) 与旧浏览器兼容(不使用过滤器) O (n ^ 2) 可选的fn回调参数，用于指定如何比较数组项

function diff(a, b, fn){ var max = Math.max(a.length, b.length); d = []; fn = typeof fn === 'function' ? fn : false for(var i=0; i < max; i++){ var ac = i < a.length ? a[i] : undefined bc = i < b.length ? b[i] : undefined; for(var k=0; k < max; k++){ ac = ac === undefined || (k < b.length && (fn ? fn(ac, b[k]) : ac == b[k])) ? undefined : ac; bc = bc === undefined || (k < a.length && (fn ? fn(bc, a[k]) : bc == a[k])) ? undefined : bc; if(ac == undefined && bc == undefined) break; } ac !== undefined && d.push(ac); bc !== undefined && d.push(bc); } return d; } alert( "Test 1: " + diff( [1, 2, 3, 4], [1, 4, 5, 6, 7] ).join(', ') + "\nTest 2: " + diff( [{id:'a',toString:function(){return this.id}},{id:'b',toString:function(){return this.id}},{id:'c',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}], [{id:'a',toString:function(){return this.id}},{id:'e',toString:function(){return this.id}},{id:'f',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}], function(a, b){ return a.id == b.id; } ).join(', ') );

这是受到了思想者接受的答案的启发，但是思想者的答案似乎假设数组是集合。如果数组是["1"，"2"]和["1"，"1"，"2"，"2"]，它就不成立了

这些数组之间的差值是["1"，"2"]。下面的解决方案是O(n*n)，所以不理想，但如果您有大数组，它也比思想者的解决方案具有内存优势。

如果你首先处理的是集合，思想者的解决方案肯定更好。如果你有一个可以访问过滤器的Javascript新版本，你也应该使用它们。这只适用于那些不处理集的人，并且正在使用旧版本的JavaScript(无论出于什么原因)…

if (!Array.prototype.diff) { 
    Array.prototype.diff = function (array) {
        // if the other array is a falsy value, return a copy of this array
        if ((!array) || (!Array.prototype.isPrototypeOf(array))) { 
            return this.slice(0);
        }

        var diff = [];
        var original = this.slice(0);

        for(var i=0; i < array.length; ++i) {
            var index = original.indexOf(array[i]);
            if (index > -1) { 
                original.splice(index, 1);
            } else { 
                diff.push(array[i]);
            }
        }

        for (var i=0; i < original.length; ++i) {
            diff.push(original[i]);
        }
        return diff;
    }
}   

下划线中的差分方法(或它的替换，Lo-Dash)也可以做到这一点:

(R)eturns the values from array that are not present in the other arrays

_.difference([1, 2, 3, 4, 5], [5, 2, 10]);
=> [1, 3, 4]

与任何下划线函数一样，你也可以以更面向对象的风格使用它:

_([1, 2, 3, 4, 5]).difference([5, 2, 10]);

转换为字符串对象类型:

[1, 1].toString() === [1, 1].toString(); // true