我正在尝试计算不包含连续字母的排列数。我的代码通过了测试，例如＆＃39; aabb＆＃39; （回答：8）和＆＃39; aab＆＃39; （回答：2），但没有通过像＃abcdefa＆＃39;（我的答案：2520;正确答案：3600）这样的案件。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');
答案 0 :(得分：3)
我认为问题是你的排列算法;你从哪里得到那个的？我尝试了另一个（在Filip Nguyen后改编自this question的答案）并按预期返回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：在回答相关问题时，我编写了一个算法，该算法不会强制所有排列，然后跳过具有相邻双精度的排列，但使用逻辑方式仅生成正确的排列。这里解释了Permutations excluding repeated characters并扩展到每个字符包含任意数量的重复：Generate all permutations of a list without adjacent equal elements
答案 1 :(得分：1)
我同意m69，这个错误似乎与你如何产生排列有关。我得到3600＆＃39; abcdefa＆＃39;通过实现不同的算法来生成排列。我的解决方案如下。由于它使用递归来生成排列，因此解决方案并不快，但如果速度不重要，您可能会发现代码更容易理解。
在排列中使用单独的函数生成数组索引值的原因是验证排列代码是否正常工作。由于输入字符串中存在重复值，因此在排列算法中调试问题更加困难。
// Simple helper function to compute all permutations of string indices
function permute_indices_helper(input) {
var result = [];
if (input.length == 0) {
return [[]];
}
for(var i = 0; i < input.length; i++) {
var head = input.splice(i, 1)[0];
var tails = permute_indices_helper(input);
for (var j = 0; j < tails.length; j++) {
tails[j].splice(0, 0, head);
result.push(tails[j]);
}
input.splice(i, 0, head); // check
}
return result;
};
// Given an array length, generate all permutations of possible indices
// for array of that length.
// Example: permute_indices(2) generates:
// [[0,1,2], [0,2,1], [1,0,2], ... , [2, 0, 1]]
function permute_indices(array_length) {
var result = [];
for (var i = 0; i < array_length; i++) {
result.push(i);
}
return permute_indices_helper(result);
}
// Rearrange letters of input string according to indices.
// Example: "car", [2, 1, 0]
// returns: "rac"
function rearrange_string(str, indices) {
var result = "";
for (var i = 0; i < indices.length; i++) {
var string_index = indices[i];
result += str[string_index];
}
return result;
}
function permAlone(str) {
var result = 0;
var permutation_indices = permute_indices(str.length);
for (var i = 0; i < permutation_indices.length; i++) {
var permuted_string = rearrange_string(str, permutation_indices[i]);
if (! permuted_string.match(/(.)\1/g)) result++;
}
return result;
}
您可以在JSFiddle上看到一个有效的例子。