var arr=[[“颜色:白色”,“颜色:黑色”],[“尺码:10”,“尺码:20”],[“材质:钻石”,“材质:水晶”,“材质:玛瑙”]];
js 写12组 组合怎么写？在线求，困扰我2天了
升级版：
var spec=[[{“spec_name”:“颜色”,“spec_type”:“1”,“spec_id”:“5”,“spec_value_id”:“11”,“spec_value_name”:“白色”},{“spec_name”:“颜色”,“spec_type”:“1”,“spec_id”:“5”,“spec_value_id”:“12”,“spec_value_name”:“黑色”}],[{“spec_name”:“尺码”,“spec_type”:“0”,“spec_id”:“6”,“spec_value_id”:“9”,“spec_value_name”:“10”},{“spec_name”:“尺码”,“spec_type”:“0”,“spec_id”:“6”,“spec_value_id”:“10”,“spec_value_name”:“20”}],[{“spec_name”:“材质”,“spec_type”:“0”,“spec_id”:“7”,“spec_value_id”:“13”,“spec_value_name”:“钻石”},{“spec_name”:“材质”,“spec_type”:“0”,“spec_id”:“7”,“spec_value_id”:“14”,“spec_value_name”:“水晶”},{“spec_name”:“材质”,“spec_type”:“0”,“spec_id”:“7”,“spec_value_id”:“15”,“spec_value_name”:“玛瑙”}]] 最终结果就是输出12种商品组合。。。。写法要支持1个数组、N个数组。。。
仅供参考, 可能有更好的算法
function getProducts(specs) {
if (!specs || specs.length == 0) {
return [];
} else {
return joinSpec([[]], specs, 0, specs.length-1);
}
function joinSpec(prevProducts, specs, i, max) {
var currentProducts = [], currentProduct, currentSpecs = specs[i];
if ( i > max ) {
return prevProducts;
}
prevProducts.forEach(function(prevProduct) {
currentSpecs.forEach(function(spec) {
currentProduct = prevProduct.slice(0);
currentProduct.push(spec);
currentProducts.push(currentProduct);
});
});
return joinSpec(currentProducts, specs, ++i, max);
}
}
var specs = [[{"spec_name":"颜色","spec_type":"1","spec_id":"5","spec_value_id":"11","spec_value_name":"白色"},{"spec_name":"颜色","spec_type":"1","spec_id":"5","spec_value_id":"12","spec_value_name":"黑色"}],[{"spec_name":"尺码","spec_type":"0","spec_id":"6","spec_value_id":"9","spec_value_name":"10"},{"spec_name":"尺码","spec_type":"0","spec_id":"6","spec_value_id":"10","spec_value_name":"20"}],[{"spec_name":"材质","spec_type":"0","spec_id":"7","spec_value_id":"13","spec_value_name":"钻石"},{"spec_name":"材质","spec_type":"0","spec_id":"7","spec_value_id":"14","spec_value_name":"水晶"},{"spec_name":"材质","spec_type":"0","spec_id":"7","spec_value_id":"15","spec_value_name":"玛瑙"}]];
console.log(getProducts(specs));
这是典型的求笛卡尔积问题，@William17的算法可以解决一般性问题，足够用了，赞一个，不过如果参数中选项较多、处理数量多的话就比较慢了，下面给出的算法用6*6数组跑了200万次的benchmark测试了一下，效率大概要快到2~3倍：
function getProducts(specs) {
specs = specs || [];
var n = specs.length;
if(n === 0) { return [] }
if(n === 1) { return specs[0] }
var save = function() {
var m = groups.length, c = [];
while(m--) { c[m] = groups[m] }
rtn.push(c);
}
var i=0, len=0, counter = [], groups = [], rtn = [];
for (; i < n; i++) {
counter[i] = 0;
groups[i] = specs[i][0];
}
save();
while(true) {
i = n - 1, len = specs[i].length;
if (++counter[i] >= len) {
while (counter[i] >= len) {
if (i === 0) { return rtn }
groups[i] = specs[i][0];
counter[i--] = 0;
counter[i]++;
len = specs[i].length;
}
}
groups[i] = specs[i][counter[i]];
save();
}
}
var specs = [[{"spec_name":"颜色","spec_type":"1","spec_id":"5","spec_value_id":"11","spec_value_name":"白色"},{"spec_name":"颜色","spec_type":"1","spec_id":"5","spec_value_id":"12","spec_value_name":"黑色"}],[{"spec_name":"尺码","spec_type":"0","spec_id":"6","spec_value_id":"9","spec_value_name":"10"},{"spec_name":"尺码","spec_type":"0","spec_id":"6","spec_value_id":"10","spec_value_name":"20"}],[{"spec_name":"材质","spec_type":"0","spec_id":"7","spec_value_id":"13","spec_value_name":"钻石"},{"spec_name":"材质","spec_type":"0","spec_id":"7","spec_value_id":"14","spec_value_name":"水晶"},{"spec_name":"材质","spec_type":"0","spec_id":"7","spec_value_id":"15","spec_value_name":"玛瑙"}]];
console.log(getProducts(specs));
