Question

我有一个函数（from Akseli Palén）来计算给定数组中X元素的组合，看起来像这样。

function k_combinations(set, k) {
  var i, j, combs, head, tailcombs;
  if (k > set.length || k <= 0) { return []; }
  if (k == set.length) { return [set]; }
  if (k == 1) {
    combs = [];
    for (i = 0; i < set.length; i++) { combs.push([set[i]]); }
    return combs;
  }
  // Assert {1 < k < set.length}
  combs = [];
  for (i = 0; i < set.length - k + 1; i++) {
    head = set.slice(i, i+1);
    tailcombs = k_combinations(set.slice(i + 1), k - 1);
    for (j = 0; j < tailcombs.length; j++) {
      combs.push(head.concat(tailcombs[j]));
    }
  }
  return combs;
}

它工作得很好，但是当给出一个大的数组和/或K值来使用它时，它会减慢到浏览器试图阻止脚本无响应的程度，所以我想知道是否有可能扩展这个函数，它接受一个调用中返回的起始位置和最大结果数？这样我可以做一些像“显示结果20-30万分之一”。有一个样本表格 http://jsbin.com/ricabomofetu/1/edit?js,output如果有人想要破解那么。

Answer 1

这样的事情怎么样？

function k_combinations(set, k, start_combo, max_results) {
    if (start_combo.length !== k)
        throw new Error("Starting combination is not of length k!");
    var cur = [];
    for (var i = 0; i < k; i++) {
        var idx = set.indexOf(start_combo[i]);
        if (idx === -1)
            throw new Error(i + "th element of starting combination isn't in the set!");
        if (i > 0 && idx <= cur[i - 1])
            throw new Error("Elements of start_combo must be in sorted order!");
        cur.push(idx);
    }

    function subset_from_indices(subset) {
        var ret = [];
        for (var i = 0; i < subset.length; i++)
            ret.push(set[subset[i]]);
        return ret;
    }

    var n = set.length;
    var results = [subset_from_indices(cur)];
    while (results.length < max_results) {
        var inc_idx = k - 1;
        while (inc_idx >= 0 && cur[inc_idx] === n - k + inc_idx)
            inc_idx--;

        if (inc_idx < 0) // no more combinations
            return results

        cur[inc_idx]++;
        for (var i = inc_idx + 1; i < k; i++) {
            cur[i] = cur[i - 1] + 1;
        }
        results.push(subset_from_indices(cur));
    }
    return results;
}

console.log(k_combinations([1, 2, 3, 4, 5], 3, [1, 3, 4], 4));
// [ [ 1, 3, 4 ], [ 1, 3, 5 ], [ 1, 4, 5 ], [ 2, 3, 4 ] ]

逐步执行K组合算法X步骤

1 个答案: