Javascript 需要帮助获得尽可能多的独特啤酒吗
比如说,我有一个酒吧,汽车在去海滩之前停下来买啤酒。每辆车都有一个行李箱大小(Javascript 需要帮助获得尽可能多的独特啤酒吗,javascript,recursion,language-agnostic,combinations,permutation,Javascript,Recursion,Language Agnostic,Combinations,Permutation,比如说,我有一个酒吧,汽车在去海滩之前停下来买啤酒。每辆车都有一个行李箱大小(remaingsum),每种啤酒都有一个大小(beer.size) 我想为客户提供啤酒组合选择(allcombination),让他们的汽车行李箱能够容纳,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,
remaingsum
),每种啤酒都有一个大小(beer.size
)
我想为客户提供啤酒组合选择(allcombination
),让他们的汽车行李箱能够容纳,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);
}
}
AllCombinations=[];
获取组合(当前组合、啤酒、剩余金额)
{
如果(剩余总和<0)
return;//总和太大;终止递归
否则{
如果(currentCombination.length>0)
{
currentCombination.sort();
var uniquePermutation=true;
对于(var i=0;i
还有另一种方法:
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;
// get all combinations (stolen from http://stackoverflow.com/questions/5752002/find-all-possible-subset-combos-in-an-array )
function combinations(array) {
return new Array(1 << array.length).fill().map(
(e1,i) => array.filter((e2, j) => i & 1 << j));
}
// filter them out if the summed sizes are > trunksize
var valids = combinations(Beers).filter(function(el) {
return el.reduce(function(a,b){return a+b.size;}, 0) <= TrunkSize;
});
console.log(valids);
let Beers=[
{id:1,大小:4},
{id:5,大小:1},
{id:10,大小:0.5},
{id:11,大小:1},
{id:12,大小:2},
{id:13,大小:1},
];
让树干大小=2;
//获取所有组合(从http://stackoverflow.com/questions/5752002/find-all-possible-subset-combos-in-an-array )
函数组合(数组){
返回新数组(1个数组.过滤器((e2,j)=>i&1个大小
var valids=组合(啤酒)。过滤器(功能(el){
return el.reduce(函数(a,b){return a+b.size;},0)下面是另一种方法:
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;
// get all combinations (stolen from http://stackoverflow.com/questions/5752002/find-all-possible-subset-combos-in-an-array )
function combinations(array) {
return new Array(1 << array.length).fill().map(
(e1,i) => array.filter((e2, j) => i & 1 << j));
}
// filter them out if the summed sizes are > trunksize
var valids = combinations(Beers).filter(function(el) {
return el.reduce(function(a,b){return a+b.size;}, 0) <= TrunkSize;
});
console.log(valids);
let Beers=[
{id:1,大小:4},
{id:5,大小:1},
{id:10,大小:0.5},
{id:11,大小:1},
{id:12,大小:2},
{id:13,大小:1},
];
让树干大小=2;
//获取所有组合(从http://stackoverflow.com/questions/5752002/find-all-possible-subset-combos-in-an-array )
函数组合(数组){
返回新数组(1个数组.过滤器((e2,j)=>i&1个大小
var valids=组合(啤酒)。过滤器(功能(el){
返回el.reduce(函数(a,b){返回a+b.size;},0)若要获得所有可能的组合而不存在重复项,可以用一组N位来表示组合,其中N=#of若要获得所有可能的组合而不存在重复项,可以用一组N位来表示组合,其中N=#of我已经编辑了您的代码,修复了使用附加数组和字符串化进行排序和检查的问题:
let Beers=[
{id:1,大小:4},
{id:5,大小:1},
{id:10,大小:0.5},
{id:11,大小:1},
{id:12,大小:2},
{id:13,大小:1},
];
让树干大小=2;
所有组合=[];
var combStrings=[]
函数GetCombinations(currentCombination、beers、remainingSum)
{
如果(剩余总和<0)
return;//总和太大;终止递归
否则{
如果(currentCombination.length>0)
{
currentCombination.sort((a,b)=>{
返回a.id>b.id
});
//var uniquePermutation=true;
var tmp=currentCombination.map((cc)=>{
返回cc.id;
})
if(combStrings.indexOf(JSON.stringify(tmp))=-1){
此.allcombines.push(当前组合);
var tmp=currentCombination.map((cc)=>{
返回cc.id;
})
push(JSON.stringify(tmp))
}
}
}
对于(变量i=0;i log(所有组合,组合字符串)
我已经编辑了您的代码,修复了使用附加数组和字符串化进行排序和检查的问题:
let Beers=[
{id:1,大小:4},
{id:5,大小:1},
{id:10,大小:0.5},
{id:11,大小:1},
{id:12,大小:2},
{id:13,大小:1},
];
让树干大小=2;
所有组合=[];
var combStrings=[]
函数GetCombinations(currentCombination、beers、remainingSum)
{
如果(剩余总和<0)
return;//总和太大;终止递归
否则{
如果(currentCombination.length>0)
{
currentCombination.sort((a,b)=>{
返回a.id>b.id
});
//var uniquePermutation=true;
var tmp=currentCombination.map((cc)=>{
返回cc.id;
})
if(combStrings.indexOf(JSON.stringify(tmp))=-1){
此.allcombines.push(当前组合);
var tmp=currentCombination.map((cc)=>{
返回cc.id;
})
push(JSON.stringify(tmp))
}
}
}
对于(变量i=0;i