javascript - 需要帮助以获取尽可能多的独特啤酒

假设我有一个酒吧，开车前往海滩前会停下来喝啤酒。每辆汽车的行李箱尺寸（remainingSum）和啤酒的尺寸（beer.size）

我想为客户提供他们的汽车后备箱适合的啤酒组合选择（AllCombinations），但组合独特。

例如，输入：

let Beers = [
    {id: 1, size: 4},
    {id: 5, size: 1},
    {id: 10, size: 0.5},
    {id: 11, size: 1},
    {id: 12, size: 2},
    {id: 13, size: 1},
];

let TrunkSize = 2;

预期产量

AllCombinations = [ // no duplicates
    [{id: 5, size: 1}, {id: 10, size: 0.5}],
    [{id: 5, size: 1}, {id: 11, size: 1}],
    [{id: 5, size: 1}, {id: 13, size: 1}],
    [{id: 10, size: 0.5}, {id: 11, size: 1}],
    [{id: 10, size: 0.5}, {id: 13, size: 1}],
    [{id: 11, size: 1}, {id: 13, size: 1}],
    [{id: 5, size: 1}],
    [{id: 11, size: 1}],
    [{id: 12, size: 2}],
    [{id: 13, size: 1}],
    [{id: 10, size: 0.5}],
]

电流输出

AllCombinations = [
    [{id: 5, size: 1}, {id: 10, size: 0.5}],  // dup a
    [{id: 5, size: 1}, {id: 11, size: 1}],    // dup c
    [{id: 5, size: 1}, {id: 13, size: 1}],    // dup d
    [{id: 10, size: 0.5}, {id: 5, size: 1}],  // dup a
    [{id: 10, size: 0.5}, {id: 11, size: 1}], // dup b
    [{id: 10, size: 0.5}, {id: 13, size: 1}], // dup e
    [{id: 11, size: 1}, {id: 13, size: 1}],   // dup f
    [{id: 11, size: 1}, {id: 10, size: 0.5}], // dup b
    [{id: 11, size: 1}, {id: 5, size: 1}],    // dup c
    [{id: 13, size: 1}, {id: 5, size: 1}],    // dup d
    [{id: 13, size: 1}, {id: 10, size: 0.5}], // dup e
    [{id: 13, size: 1}, {id: 11, size: 1}],   // dup f
    [{id: 5, size: 1}],
    [{id: 11, size: 1}],
    [{id: 12, size: 2}],
    [{id: 13, size: 1}],
    [{id: 10, size: 0.5}]
]

当前功能：

AllCombinations = [];
GetCombinations(currentCombination, beers, remainingSum)
{

    if (remainingSum < 0)
        return;// Sum is too large; terminate recursion

    else {
        if (currentCombination.length > 0)
        {
            currentCombination.sort();
            var uniquePermutation = true;

            for (var i = 0; i < this.AllCombinations.length; i++)
            {
                if (currentCombination.length == this.AllCombinations[i].length)
                {
                    for (var j = 0; currentCombination[j] == this.AllCombinations[i][j] && j < this.AllCombinations[i].length; j++);  // Pass

                    if (j == currentCombination.length) {
                        uniquePermutation = false;
                        break;
                    }
                }
            }

            if (uniquePermutation)
                this.AllCombinations.push(currentCombination);
        }
    }

    for (var i = 0; i < beers.length; i++) {
        var newChoices = beers.slice();
        var newCombination = currentCombination.concat(newChoices.splice(i, 1));
        var newRemainingSum = remainingSum - beers[i].size;
        this.GetCombinations(newCombination, newChoices, newRemainingSum);
    }
}

最佳答案

我已经编辑了您的代码，修复了排序和其他数组检查以及stringify的检查：

let Beers = [
    {id: 1, size: 4},
    {id: 5, size: 1},
    {id: 10, size: 0.5},
    {id: 11, size: 1},
    {id: 12, size: 2},
    {id: 13, size: 1},
];

let TrunkSize = 2;

AllCombinations = [];
var combStrings = []
function GetCombinations(currentCombination, beers, remainingSum)
{

    if (remainingSum < 0)
        return;// Sum is too large; terminate recursion

    else {
        if (currentCombination.length > 0)
        {
            currentCombination.sort((a,b)=>{
              return a.id > b.id
            });
            //var uniquePermutation = true;

            var tmp = currentCombination.map((cc)=>{
                  return cc.id;
               })

            if (combStrings.indexOf(JSON.stringify(tmp)) == -1) {
                this.AllCombinations.push(currentCombination);
                var tmp = currentCombination.map((cc)=>{
                  return cc.id;
                })
                combStrings.push(JSON.stringify(tmp))
            }
        }
    }

    for (var i = 0; i < beers.length; i++) {
        var newChoices = beers.slice();
        var newCombination = currentCombination.concat(newChoices.splice(i, 1));
        var newRemainingSum = remainingSum - beers[i].size;
        this.GetCombinations(newCombination, newChoices, newRemainingSum);
    }
}
GetCombinations([],Beers,TrunkSize)
console.log(AllCombinations,combStrings)