JavaScript 排列

Question

我正在尝试计算不包含连续字母的排列数。我的代码通过了像“aabb”(答案:8)和“aab”(答案:2)这样的测试，但没有通过像“abcdefa”这样的情况(我的答案:2520；正确答案:3600)。这是我的代码:

function permAlone(str) {

    var totalPerm = 1;
    var result = [];

    //assign the first letter
    for (var i = 0; i < str.length; i++) {
        var firstNum = str[i];
        var perm = firstNum;

        //create an array from the remaining letters in the string
        for (var k = 0; k < str.length; k++) {
            if (k !== i) {
                perm += str[k];
            }
        }

        //Permutations: get the last letter and change its position by -1;
        //Keep changing that letters's position by -1 until its index is 1;
        //Then, take the last letter again and do the same thing;
        //Keep doing the same thing until the total num of permutations of the number of items in the string -1 is reached (factorial of the number of items in the string -1 because we already established what the very first letter must be).

        var permArr = perm.split("");
        var j = permArr.length - 1;
        var patternsLeft = totalNumPatterns(perm.length - 1);

        while (patternsLeft > 0) {
            var to = j - 1;
            var subRes = permArr.move(j, to);
            console.log(subRes);

            if (noDoubleLettersPresent(subRes)) {
                result.push([subRes]);
            }

            j -= 1;
            if (j == 1) {
                j = perm.length - 1;
            }
            patternsLeft--;
        }
    }
    return result.length;
}

Array.prototype.move = function(from, to) {
    this.splice(to, 0, (this.splice(from, 1))[0]);
    return this.join("");
};

function totalNumPatterns(numOfRotatingItems) {
    var iter = 1;
    for (var q = numOfRotatingItems; q > 1; q--) {
        iter *= q;
    }
    return iter;
}

function noDoubleLettersPresent(str) {
    if (str.match(/(.)\1/g)) {
        return false;
    } else {
        return true;
    }
}

permAlone('abcdefa');

Answer 1

我认为问题出在你的排列算法上；你从哪里得到那个的？我用另一个尝试了它(在 Filip Nguyen 之后，改编自他对 this question 的回答)，它按预期返回 3600。

function permAlone(str) {
    var result = 0;
    var fact = [1];
    for (var i = 1; i <= str.length; i++) {
        fact[i] = i * fact[i - 1];
    }
    for (var i = 0; i < fact[str.length]; i++) {
        var perm = "";
        var temp = str;
        var code = i;
        for (var pos = str.length; pos > 0; pos--) {
            var sel = code / fact[pos - 1];
            perm += temp.charAt(sel);
            code = code % fact[pos - 1];
            temp = temp.substring(0, sel) + temp.substring(sel + 1);
        }
        console.log(perm);
        if (! perm.match(/(.)\1/g)) result++;
    }
    return result;
}

alert(permAlone('abcdefa'));

更新:为了回答一个相关问题，我编写了一个算法，它不仅强制所有排列，然后跳过具有相邻 double 的排列，而且使用逻辑方法仅生成正确的排列。解释如下:Permutations excluding repeated characters并扩展为包括每个字符的任意数量的重复:Generate all permutations of a list without adjacent equal elements