我明白了，问题在于复杂性是平方的.有一个技巧可以做到，只需使用关联数组"即可.

I see, the problem there is that the complexity is squared. There is one trick to do it, it's simply by using "Associative arrays".

您可以获取数组，对其进行循环，然后将数组的值作为键添加到关联数组中.由于它不允许重复键，您将自动删除重复项.

You can get the array, loop over it, and add the value of the array as a key to the associative array. Since it doesn't allow duplicated keys, you will automatically get rid of the duplicates.

由于您在比较时正在寻找标题和艺术家，您实际上可以尝试使用以下内容:

Since you are looking for title and artist when comparing, you can actually try to use something like:

var arrResult = {};
for (i = 0, n = arr.length; i < n; i++) {
    var item = arr[i];
    arrResult[ item.title + " - " + item.artist ] = item;
}

然后您只需再次循环 arrResult，并重新创建数组.

Then you just loop the arrResult again, and recreate the array.

var i = 0;
var nonDuplicatedArray = [];    
for(var item in arrResult) {
    nonDuplicatedArray[i++] = arrResult[item];
}

更新以包含保罗的评论.谢谢！