最简单的代码来检查数组可能的组合

Question

在javascript中实现get array可能组合的最简单，无库的代码是什么？我想写

var preDefinedCombinations = [
    ['a', 'b', 'c'],
    ['a', 'c', 'd'],
    ['a'],
    ['b'],
    ['c']
];
var selectedvalues = ['a', 'b', 'd'];
getPossbileCombination(selectedvalues, preDefinedCombinations);

并获取

 ['a','b','c'],['a','c','d']

我尝试了下面的代码，但没有按预期工作，如果用户选择'a'，'b'而不是警报显示他选择'c'进行有效组合，它总是给我错过的术语。但如果用户选择'a'，'b'和amp; ['d']它不会为所选值提供可能的组合。

var preDefinedCombinations = [['a','b','c'], ['a','c','d'], ['a'], ['b'], ['c']];
var selectedvalues = ['a','b','d'];

function selectOption(value){
   var checkIndex = selectedvalues.indexof(value);
   if(checkIndex == -1){
       selectedvalues.push(value);
   }else{
       selectedvalues.splice(checkIndex, -1);
   }

}

function intersect_safe(a, b)
{
 var ai=0, bi=0;
 var result = new Array();

 while( ai < a.length && bi < b.length )
 {
    if      (a[ai] < b[bi] ){ ai++; }
    else if (a[ai] > b[bi] ){ bi++; }
    else /* they're equal */
    {
      result.push(a[ai]);
      ai++;
      bi++;
    }
 }

 return result;
}

function checkVaildCombination(){
   if(preDefinedCombinations.indexOf(selectedvalues) == -1){
       var missing_bits = [];
       for (var i = 0; i < preDefinedCombinations.length; i++) {
           var missing_bit = intersect_safe(preDefinedCombinations[i], selectedvalues);
           if (missing_bit.length == selectedvalues.length) { // candidate for valid answer
               missing_bit = [];
               for (var j = 0; j < preDefinedCombinations[i].length; j++) {
                   if (selectedvalues.indexOf(preDefinedCombinations[i][j]) == -1) { // actual missing bit
                       missing_bit.push(preDefinedCombinations[i][j]);
                   }
               }

               missing_bits.push(missing_bit);
           }
       }

       alert('Invalid Combination, if you select any of these you`ll get a valid combination:\n' + missing_bits.toString());
   }else{
       alert('Valid Combination');
   }
}