我有麻烦使这项工作。我对这个算法的理解（基于维基百科的解释）是它应该像我在下面的代码中那样工作。但有些事情并没有奏效。如果有2,3,4个数字或字母，一切都运行良好，但如果有更多的数字或字母，算法不起作用。
这里有什么问题吗？
以下是代码：
//swap two letters at the given index
String.prototype.swapLetters = function(i){
var a = this.split("");
var temp = this[i];
a[i]=a[i+1];
a[i+1]=temp;
return a.join("");
}
//get all posible combinations for given string
function combs(n){
if(n === 0){
return 1;
} else {
return n*(combs(n-1));
}
}
function permAlone(str) {
var reg = new RegExp(/(\w)\1+/);
var c = 0;
//the point of this function is to implement Steinhaus–Johnson–Trotter algorithm
// to list all permutations. The algorithm should work like this: if we have 1234
// 1. 2134; -> 2. 2314; -> 3. 2341;
// now is break switch first two letters so:
// 4. 3241;
// then proced is oposite direction
// 5. 3214; -> 3124;...
function permutations(str, p, n, b, perms){
var index = p-n;
var len = str.length-2;
console.log(str);
if(perms===0){
return;
}
//if p is less then or equal and b is zStart then increase p by 1
//if p is equal to len do not increment p just call function with b===zBreak (call function to swich first two letters)
if(p <= len && b==="zStart"){
if(p === len){
b = "zBreak";
} else {
p = p+1;
}
return permutations(str.swapLetters(index), p, n, b, perms-1);
}
//if n is less then or equal and b is lStart then increase n by 1
//if n is equal to len do not increment n just call function with b===lBreak (call function to swich first last two letters)
if (n <= len && b==="lStart"){
if(n === len){
b = "lBreak";
} else {
n = n+1;
}
return permutations(str.swapLetters(index), p, n, b, perms-1);
}
//if b is zBreak swap first two letters and set b to be lStart because
//next break should swap last two letters
if(b==="zBreak"){
return permutations(str.swapLetters(0), p, n, "lStart", perms-1);
} else if(b==="lBreak"){
return permutations(str.swapLetters(len), 0, 0, "zStart", perms-1);
}
return permutations(str.swapLetters(index), p, n, b, perms-1);
}
permutations(str, 0, 0, "zStart", combs(str.length));
}
permAlone('1234');