我有两个JavaScript数组:

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

我希望输出为:

var array3 = ["Vijendra","Singh","Shakya"];

输出数组应删除重复的单词。

如何在JavaScript中合并两个数组,以便从每个数组中只获得唯一的项目,其顺序与它们插入原始数组的顺序相同?


当前回答

var arr1 = [1, 3, 5, 6];
var arr2 = [3, 6, 10, 11, 12];
arr1.concat(arr2.filter(ele => !arr1.includes(ele)));
console.log(arr1);

output :- [1, 3, 5, 6, 10, 11, 12]

其他回答

新解决方案(使用Array.prototype.indexOf和Array.prototype.cocat):

Array.prototype.uniqueMerge = function( a ) {
    for ( var nonDuplicates = [], i = 0, l = a.length; i<l; ++i ) {
        if ( this.indexOf( a[i] ) === -1 ) {
            nonDuplicates.push( a[i] );
        }
    }
    return this.concat( nonDuplicates )
};

用法:

>>> ['Vijendra', 'Singh'].uniqueMerge(['Singh', 'Shakya'])
["Vijendra", "Singh", "Shakya"]

Array.prototype.indexOf(用于internet explorer):

Array.prototype.indexOf = Array.prototype.indexOf || function(elt)
  {
    var len = this.length >>> 0;

    var from = Number(arguments[1]) || 0;
    from = (from < 0) ? Math.ceil(from): Math.floor(from); 
    if (from < 0)from += len;

    for (; from < len; from++)
    {
      if (from in this && this[from] === elt)return from;
    }
    return -1;
  };

只需避开嵌套循环(O(n^2))和.indexOf()(+O(n))。

函数合并(a,b){var哈希={};变量i;对于(i=0;i<a.length;i++){hash[a[i]=真;}对于(i=0;i<b.length;i++){hash[b[i]]=真;}return Object.keys(哈希);}var array1=[“Vijendra”,“Singh”];var array2=[“Singh”,“Shakya”];var array3=合并(array1,array2);console.log(array3);

对于大型输入,更好的选择是对数组进行排序。然后合并它们。

function sortFunction(a, b) {
        return a - b;
}

arr1.sort(sortFunction);
arr2.sort(sortFunction);

function mergeDedup(arr1, arr2) {

    var i = 0, j = 0, result = [];
    while (i < arr1.length && j < arr2.length) {

        if (arr1[i] < arr2[j]) {
            writeIfNotSameAsBefore(result, arr1[i]);
            i++;
        }
        else if (arr1[i] > arr2[j]) {
            writeIfNotSameAsBefore(result, arr2[j]);
            j++;
        }
        else {
            writeIfNotSameAsBefore(result, arr1[i]);
            i++;
            j++;
        }

    }

    while (i < arr1.length) {
        writeIfNotSameAsBefore(result, arr1[i]);
        i++;
    }

    while (j < arr2.length) {
        writeIfNotSameAsBefore(result, arr2[j]);
        j++;
    }
    return result;
}

function writeIfNotSameAsBefore(arr, item) {
    if (arr[arr.length - 1] !== item) {
        arr[arr.length] = item;
    }
    return arr.length;
}

排序将采用O(nlogn+mlogm),其中n和m是数组的长度,O(x)用于合并,其中x=Max(n,m);

这很简单,可以用jQuery在一行中完成:

var arr1 = ['Vijendra', 'Singh'], arr2 =['Singh', 'Shakya'];

$.unique(arr1.concat(arr2))//one line

["Vijendra", "Singh", "Shakya"]
Array.prototype.union = function (other_array) {
/* you can include a test to check whether other_array really is an array */
  other_array.forEach(function(v) { if(this.indexOf(v) === -1) {this.push(v);}}, this);    
}