JS计算两个数组的交集、差集、并集、补集(多种实现方式)
方法一:最普遍的做法1、直接使用 filter、concat 来计算2、对 Array 进行扩展方法二:使用 ES6 语法实现方法三:使用 jQuery 实现运行结果方法一:最普遍的做法
使用 ES5 语法来实现虽然会麻烦些,但兼容性最好,不用考虑浏览器 JavaScript 版本。也不用引入其他第三方库。
1、直接使用 filter、concat 来计算
var a = [1,2,3,4,5]var b = [2,4,6,8,10]//交集var intersect = a.filter(function(v){return b.indexOf(v) > -1 })//差集var minus = a.filter(function(v){return b.indexOf(v) == -1 })//补集var complement = a.filter(function(v){return !(b.indexOf(v) > -1) }).concat(b.filter(function(v){return !(a.indexOf(v) > -1)}))//并集var unionSet = a.concat(b.filter(function(v){return !(a.indexOf(v) > -1)}));console.log("数组a:", a);console.log("数组b:", b);console.log("a与b的交集:", intersect);console.log("a与b的差集:", minus);console.log("a与b的补集:", complement);console.log("a与b的并集:", unionSet);
2、对 Array 进行扩展
(1)为方便使用,我们可以对数组功能进行扩展,增加一些常用的方法。
//数组功能扩展//数组迭代函数Array.prototype.each = function(fn){fn = fn || Function.K;var a = [];var args = Array.prototype.slice.call(arguments, 1);for(var i = 0; i < this.length; i++){var res = fn.apply(this,[this[i],i].concat(args));if(res != null) a.push(res);}return a;};//数组是否包含指定元素Array.prototype.contains = function(suArr){for(var i = 0; i < this.length; i ++){if(this[i] == suArr){return true;}}return false;}//不重复元素构成的数组Array.prototype.uniquelize = function(){var ra = new Array();for(var i = 0; i < this.length; i ++){if(!ra.contains(this[i])){ra.push(this[i]);}}return ra;};//两个数组的交集Array.intersect = function(a, b){return a.uniquelize().each(function(o){return b.contains(o) ? o : null});};//两个数组的差集Array.minus = function(a, b){return a.uniquelize().each(function(o){return b.contains(o) ? null : o});};//两个数组的补集plement = function(a, b){return Array.minus(Array.union(a, b),Array.intersect(a, b));};//两个数组并集Array.union = function(a, b){return a.concat(b).uniquelize();};
(2)使用样例
var a = [1,2,3,4,5]var b = [2,4,6,8,10]console.log("数组a:", a);console.log("数组b:", b);console.log("a与b的交集:", Array.intersect(a, b));console.log("a与b的差集:", Array.minus(a, b));console.log("a与b的补集:", plement(a, b));console.log("a与b的并集:", Array.union(a, b));
方法二:使用 ES6 语法实现
(1)而在 ES6 中我们可以借助扩展运算符(…)以及 Set 的特性实现相关计算,代码也会更加简单些。
var a = [1,2,3,4,5]var b = [2,4,6,8,10]console.log("数组a:", a);console.log("数组b:", b);var sa = new Set(a);var sb = new Set(b);// 交集let intersect = a.filter(x => sb.has(x));// 差集let minus = a.filter(x => !sb.has(x));// 补集let complement = [...a.filter(x => !sb.has(x)), ...b.filter(x => !sa.has(x))];// 并集let unionSet = Array.from(new Set([...a, ...b]));console.log("a与b的交集:", intersect);console.log("a与b的差集:", minus);console.log("a与b的补集:", complement);console.log("a与b的并集:", unionSet);
(2)借助includes以及箭头函数,代码也会更加简单些。
var a = [1,2,3,4,5];var b = [2,4,6,8,10];//交集var intersect = a.filter(v => b.includes(v));//差集var minus = a.filter(v => !b.includes(v));//补集var complement = a.filter(v => !b.includes(v)).concat(b.filter(v => !a.includes(v)))//并集var unionSet = a.concat(b.filter(v => !a.includes(v)));console.log("a与b的交集:", intersect);console.log("a与b的差集:", minus);console.log("a与b的补集:", complement);console.log("a与b的并集:", unionSet);
方法三:使用 jQuery 实现
引入 jQuery,那么实现起来也很简单。
var a = [1,2,3,4,5]var b = [2,4,6,8,10]console.log("数组a:", a);console.log("数组b:", b);// 交集let intersect = $(a).filter(b).toArray();// 差集let minus = $(a).not(b).toArray();// 补集let complement = $(a).not(b).toArray().concat($(b).not(a).toArray());// 并集let unionSet = $.unique(a.concat(b));console.log("a与b的交集:", intersect);console.log("a与b的差集:", minus);console.log("a与b的补集:", complement);console.log("a与b的并集:", unionSet);