一个所有元素都必须用到的子集组合问题？

用javascript写了一下，改java也容易，

const arr=['A','B','C']


//获取所有子集
function generateSubsets(arr, subset = [[]]) {
  if (arr.length === 0) {
    return subset;
  } else {
    const current = arr[0];
    const newSubset = [];
    subset.forEach(sub => {
      newSubset.push(sub.concat(current), sub);
    });
    return generateSubsets(arr.slice(1), newSubset);
  }
}


//取子集一半项的差集，因为[[A],[B,C]]认为等同于[[B,C],[A]]
//返回子集和差集的，数组就是题目要求的结果
//单独处理每个单项结果
function generateDiffSets(arr,b){
  
  var result=[]
  for(var i=0;i<b.length/2;i++){
    //console.log("ddd",arr)
    var diffs = arr.filter(function(v){ 
      return b[i].indexOf(v) == -1
    })
    //console.log(b[i],diffs)
    result.push([b[i],diffs])
  }
  
  //添加单项结果
  var t = arr.map((i)=>{
    return [i]
  })
  
  result.push(t)
  
  return result
}



var subsets= generateSubsets(arr)

var results =generateDiffSets(arr,subsets)

console.log(results)