No repeats please (Freecodecamp Advanced algorithm 6)

Rearrange the characters in a string to generate a new string that returns the number of strings in the newly generated string that do not have consecutive repeating characters. Continuous repetition only takes a single character

For example, aab you should return 2 because it has a total of 6 permutations (,,,, aab aab aba aba baa , baa ), but only two (and aba aba ) have no consecutive repeating characters (in this case a ).

function Permalone (str) {//use regular match to repeat var regex =/(.)                    \1+/g;//brackets represent groupings, \1 means getting exactly the same content as the first group//converting the string to an array for easy handling of var arr = Str.split (");                    Stores a full array of var permutations = [];                    var tmp;                        function function: Swap the location function swap (index1, index2) {tmp = ARR[INDEX1];                        ARR[INDEX1] = Arr[index2];                    ARR[INDEX2] = tmp; }//Use heap ' s algorithm to generate a fully arranged array function generate (n) {if (n = =                        = 1) {Permutations.push (Arr.join (")); } else {for (var i = 0;i<N; ++i) {generate (n-1);                            Swap (n% 2 0:i, n-1);                    }}} generate (Arr.length); Filter out arrays of non-sequential repeats var filtered= Permutations.filter (function (String){return!string.match (regex);                    });                Return the number of return filtered.length; } Alert (Permalone ("AABFCDF"));

Heap ' s algorithm full permutation algorithm

procedure Generate (N:integer, A:array of any):    if n = 1 then        output (A)    else        <   -1; i +do            generate (n-1, A)            if n was even then                swap (A[i], a[n-1])            Else
    swap (A[0], a[n-1])            End If         end for         generate (n-1, A)    End If

When I=n, the algorithm produces all the full permutations;

If n is odd, the first element in the permutation is swapped with the last element. If n is an even number, the first element of the arrangement is interchanged with the last element.

No repeats please (Freecodecamp Advanced algorithm 6)