高效的算法来获取对象中所有项的组合 [英] Efficient algorithm to get the combinations of all items in object
问题描述
给定一个带有n个键的数组或对象,我需要找到长度 x
的所有组合。
给定 X
是可变的。 binomial_coefficient(n,x)
。
Given an array or object with n keys, I need to find all combinations with length x
.
Given X
is variable. binomial_coefficient(n,x)
.
目前我使用的是:
function combine(items) {
var result = [];
var f = function(prefix, items) {
for (var i = 0; i < items.length; i++) {
result.push(prefix + items[i]);
f(prefix + items[i], items.slice(i + 1));
}
}
f('', items);
return result;
}
var combinations = combine(["a", "b", "c", "d"]);
输出为:
["a", "ab", "abc", "abcd", "abd", "ac", "acd", "ad", "b", "bc", "bcd", "bd", "c", "cd", "d"]
所以如果我希望二项式系数 x = 3
从 n = 4
我选择长度等于3的所有字符串。 {abc,abd,acd,bcd}。
So if I want the binomial coefficient x=3
from n=4
I select all the strings with length equal to three. {abc, abd, acd, bcd}.
所以我分两步完成。
有吗一个更有效的算法,复杂度更低?
Is there a more efficient algorithm with smaller complexity?
链接: 解决方案性能(JSPerf)
推荐答案
你的算法几乎是 O(2 ^ n)
,你可以丢弃很多组合,但元素的数量将是(n!*(nx)!)/ x!
。
Your algorithm is almost O(2^n)
, you can discard a lot of combinations, but the num of elements will be (n! * (n-x)!) / x!
.
要丢弃无用的组合,你可以使用索引数组。
To discard the useless combinations you can use an indexed array.
function combine(items, numSubItems) {
var result = [];
var indexes = new Array(numSubItems);
for (var i = 0 ; i < numSubItems; i++) {
indexes[i] = i;
}
while (indexes[0] < (items.length - numSubItems + 1)) {
var v = [];
for (var i = 0 ; i < numSubItems; i++) {
v.push(items[indexes[i]]);
}
result.push(v);
indexes[numSubItems - 1]++;
var l = numSubItems - 1; // reference always is the last position at beginning
while ( (indexes[numSubItems - 1] >= items.length) && (indexes[0] < items.length - numSubItems + 1)) {
l--; // the last position is reached
indexes[l]++;
for (var i = l +1 ; i < numSubItems; i++) {
indexes[i] = indexes[l] + (i - l);
}
}
}
return result;
}
var combinations = combine(["a", "b", "c", "d"], 3);
console.log(JSON.stringify(combinations));
For例如,第一个组合具有索引: [0,1,2]
和元素 [a,b,c ]
。要计算下一个组合,它得到最后一个索引 2
并尝试增加,如果增量低于最大位置(在这种情况下 4)
),达到下一个组合,但如果不是,则必须增加到之前的索引。
For example, the first combination have the indexes: [0, 1, 2]
and the elements ["a", "b", "c"]
. To compute the next combination, It get the last index 2
and try to increment, if the increment is lower than the max position (in this case 4
), the next combination is reached, but if It is not, It must increment to a previous index.
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