本文介绍了JavaScript中多个数组的笛卡尔积的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
举例来说，
cartesian（[1,2]，[10,20]，[100,200,300]）//应该是 // [[1,10,100]，[1,10,200]， [1,10,300]，[2,10,100]，[2,10,200] ...]解决方案
使用 reduce 和解决问题的功能性解决方案（不含可变变量！） c $ c> flatten ，由 underscore.js 提供：
function cartesianProductOf（）{ return _.reduce（arguments，function（a，b）{ return _.flatten（_。map（a，function（x）{ return _.map（b，function（y）{ return x.concat（[y]）; }）; }），true）; }，[[]]）; }; cartesianProductOf（[1，2]，[3，4]，['a'，'b']）; // [[1,3，a]，[1,3，b]，[1,4，a]，[1,4，b]，[2,3，a ]，[2,3，b]，[2,4，a]，[2,4，b]]备注：此解决方案的灵感来源于 cwestblog/2011/05/02/cartesian-product-of-multiple-arrays/
How would you implement the Cartesian product of multiple arrays in JavaScript?
As an example,
cartesian([1,2],[10,20],[100,200,300]) //should be // [[1,10,100],[1,10,200],[1,10,300],[2,10,100],[2,10,200]...]解决方案
Here is a functional solution to the problem (without any mutable variable!) using reduce and flatten, provided by underscore.js:
function cartesianProductOf() { return _.reduce(arguments, function(a, b) { return _.flatten(_.map(a, function(x) { return _.map(b, function(y) { return x.concat([y]); }); }), true); }, [ [] ]); }; cartesianProductOf([1, 2], [3, 4], ['a', 'b']); // [[1,3,"a"],[1,3,"b"],[1,4,"a"],[1,4,"b"],[2,3,"a"],[2,3,"b"],[2,4,"a"],[2,4,"b"]]Remark: This solution was inspired by cwestblog/2011/05/02/cartesian-product-of-multiple-arrays/
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JavaScript中多个数组的笛卡尔积
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