以下是带有对象数组的对象的选项：

const a = [{param1: "1", param2: 1},{param1: "2", param2: 2},{param1: "4", param2: 4}]
const b = [{param1: "1", param2: 1},{param1: "4", param2: 5}]


var result = a.concat(b.filter(item =>
         !JSON.stringify(a).includes(JSON.stringify(item))
    ));

console.log(result);
//Result [{param1: "1", param2: 1},{param1: "2", param2: 2},{param1: "4", param2: 4},{param1: "4", param2: 5}]

这是我的解决方案https://gist.github.com/4692150深度相等且易于使用的结果：

function merge_arrays(arr1,arr2)
{
   ... 
   return {first:firstPart,common:commonString,second:secondPart,full:finalString}; 
}

console.log(merge_arrays(
[
[1,"10:55"] ,
[2,"10:55"] ,
[3,"10:55"]
],[
[3,"10:55"] ,
[4,"10:55"] ,
[5,"10:55"]
]).second);

result:
[
[4,"10:55"] ,
[5,"10:55"]
]

如果要检查唯一对象，请在比较中使用JSON.stringify。

function arrayUnique(array) {
    var a = array.concat();
    for(var i=0; i<a.length; ++i) {
        for(var j=i+1; j<a.length; ++j) {
            if(JSON.stringify(a[i]) === JSON.stringify(a[j]))
                a.splice(j--, 1);
        }
    }

    return a;
}

Array.prototype.add = function(b){
    var a = this.concat();                // clone current object
    if(!b.push || !b.length) return a;    // if b is not an array, or empty, then return a unchanged
    if(!a.length) return b.concat();      // if original is empty, return b

    // go through all the elements of b
    for(var i = 0; i < b.length; i++){
        // if b's value is not in a, then add it
        if(a.indexOf(b[i]) == -1) a.push(b[i]);
    }
    return a;
}

// Example:
console.log([1,2,3].add([3, 4, 5])); // will output [1, 2, 3, 4, 5]

我的一便士半：

Array.prototype.concat_n_dedupe = function(other_array) {
  return this
    .concat(other_array) // add second
    .reduce(function(uniques, item) { // dedupe all
      if (uniques.indexOf(item) == -1) {
        uniques.push(item);
      }
      return uniques;
    }, []);
};

var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];

var result = array1.concat_n_dedupe(array2);

console.log(result);

如果像我一样，您需要支持较旧的浏览器，这适用于IE6+

function es3Merge(a, b) {
    var hash = {},
        i = (a = a.slice(0)).length,
        e;

    while (i--) {
        hash[a[i]] = 1;
    }

    for (i = 0; i < b.length; i++) {
        hash[e = b[i]] || a.push(e);
    }

    return a;
};

http://jsperf.com/merge-two-arrays-keeping-only-unique-values/22