我的思路是这样的:

先看成一个组合的问题([x]选n)

再写一个过滤器把含有有同一组元素的组合过滤掉.

这样扩展性会比较好;

//合并数组

function concat(...arrs){

return [].concat(...arrs);

}

//过滤器 判断是不符合规则

function getFilter(...arrs){

return function(arr){

var haveArr = new Set();

for(let v of arr){

let perr = arrs.find((a)=>{

return a.includes(v)

})

if(haveArr.has(perr)){

return false

}else{

haveArr.add(perr)

}

}

return true;

}

}

//递归组合实现

function zuhe(arr,n){

var pass = new Set();

var zh = []

if(n<0 || arr.length < n){throw '参数错误'}

if(n == 1){

for(let v of arr){

if(!pass.has(v)){ //去重

pass.add(v);

zh.push([v]);

}

}

return zh;

}else if(arr.length == n){

return [arr];

}

for(let i=0; i <= arr.length-n; i++ ){

let v = arr[i];

if(!pass.has(v)){ //去重

pass.add(v);

let _zh = zuhe(arr.slice(i+1),n-1).map(a=>{

a.unshift(v)

return a;

})

zh = zh.concat(_zh)

}

}

return zh;

}

//----------以上是所需要的工具函数

var arr1 = ["a","b"];

var arr2 = ["A","B"];

var arr3 = ["1","2"];

var allArr = concat(arr1,arr2,arr3); //合并数组

var filter = getFilter(arr1,arr2,arr3); //生成过滤器

var zh1 = zuhe(allArr,2).filter(v=>filter(v)).map(v=>v.join("_")); // x选2>过滤>组装成需要格式

var zh2 = zuhe(allArr,3).filter(v=>filter(v)).map(v=>v.join("_")); // x选3>过滤后>组装成需要格式

console.log(zh1);

console.log(zh2);

/*

["a_A","a_B","a_1","a_2","b_A","b_B","b_1","b_2","A_1","A_2","B_1","B_2"]

["a_A_1","a_A_2","a_B_1","a_B_2","b_A_1","b_A_2","b_B_1","b_B_2"]

*/

---------------- 补充

想明白 其实还是一个组合的问题 只是组合的结果要处理一下 稍微改一下 zuhe 函数:

function zuhe(arr,n){

var pass = new Set();

var zh = []

if(n<0 || arr.length < n){throw '参数错误'}

if(n == 1){

for(let _arr of arr){

for(let v of _arr){

if(!pass.has(v)){ //去重

pass.add(v);

zh.push([v]);

}

}

}

return zh;

}

for(let i=0; i <= arr.length-n; i++ ){

for(let v of arr[i]){

if(!pass.has(v)){ //去重

pass.add(v);

let _zh = zuhe(arr.slice(i+1),n-1).map(a=>{

a.unshift(v)

return a;

})

zh = zh.concat(_zh)

}

}

}

return zh;

}

var arr1 = ["a","b","c"];

var arr2 = ["A","B"];

var arr3 = ["1","2"];

var zh2 = zuhe([arr1,arr2,arr3],2)

var zh3 = zuhe([arr1,arr2,arr3],3)

console.log(zh2)

console.log(zh3)