var a = [1,2,3,4,5]var b = [2,4,6,8,10]//交集var c = a.filter(function(v){ return b.indexOf(v) > -1 })//差集var d = a.filter(function(v){ return b.indexOf(v) == -1 })//补集var e = a.filter(function(v){ return !(b.indexOf(v) > -1) }) .concat(b.filter(function(v){ return !(a.indexOf(v) > -1)}))//并集var f = 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:", c);console.log("a与b的差集:", d);console.log("a与b的补集:", e);console.log("a与b的并集:", f); |
//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Array.complement = 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();}; |
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:", Array.complement(a, b));console.log("a与b的并集:", Array.union(a, b)); |
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); |
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); |
JS - 计算两个数组的交集、差集、并集、补集(多种实现方式)
原文:https://www.cnblogs.com/xuey/p/11448762.html