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2012-10-07T17:08:12.950
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Method to generate all permuatuons of spinnable strings I've implemented a simple method to solve this problem. It takes an ArrayList argument containing spinnable text string(s). I use it to generate all the permutations of multiple spinnable strings. It comes with extra functionality of support of optional blocks, surronded by "[ ]" brackets. Eq.: If you have a single string object in the ArrayList with content of: {A | {B1 | B2 } [B optional] } It populates the array list with all the permutations, "extracted" Contents after invocation of method: A B1 B1 B optional B2 B2 B optional You can also pass multiple strings as argument to generate permutations for all of them: Eg.: Input: ArraList with two string {A1 | A2} {B1 | B2} Contents after invocation: A1 A2 B1 B2 This implementation works by always finding the inner most bracket pair in the first spinnable section, then extract it. I do this until all the special {}, [] characters are removed. <pre><code>private void ExtractVersions(ArrayList list) { ArrayList IndicesToRemove = new ArrayList(); for (int i = 0; i < list.Count; i++) { string s = list[i].ToString(); int firstIndexOfCurlyClosing = s.IndexOf('}'); int firstIndexOfBracketClosing = s.IndexOf(']'); if ((firstIndexOfCurlyClosing > -1) || (firstIndexOfBracketClosing > -1)) { char type = ' '; int endi = -1; int starti = -1; if ((firstIndexOfBracketClosing == -1) && (firstIndexOfCurlyClosing > -1)) { // Only Curly endi = firstIndexOfCurlyClosing; type = '{'; } else { if ((firstIndexOfBracketClosing > -1) && (firstIndexOfCurlyClosing == -1)) { // Only bracket endi = firstIndexOfBracketClosing; type = '['; } else { // Both endi = Math.Min(firstIndexOfBracketClosing, firstIndexOfCurlyClosing); type = s[endi]; if (type == ']') { type = '['; } else { type = '{'; } } } starti = s.Substring(0, endi).LastIndexOf(type); if (starti == -1) { throw new Exception("Brackets are not valid."); } // start index, end index and type found. -> make changes if (type == '[') { // Add two new lines, one with the optional part, one without it list.Add(s.Remove(starti, endi - starti+1)); list.Add(s.Remove(starti, 1).Remove(endi-1, 1)); IndicesToRemove.Add(i); } else if (type == '{') { // Add as many new lines as many alternatives there are. This must be an in most bracket. string alternatives = s.Substring(starti + 1, endi - starti - 1); foreach(string alt in alternatives.Split('|')) { list.Add(s.Remove(starti,endi-starti+1).Insert(starti,alt)); } IndicesToRemove.Add(i); } } // End of if( >-1 && >-1) } // End of for loop for (int i = IndicesToRemove.Count-1; i >= 0; i--) { list.RemoveAt((int)IndicesToRemove[i]); } } </code></pre> I hope I've helped. Maybe it is not the simplest and best implementation, but it works well for me. Please feedback, and vote!
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1. POHow can find all permutations of spinning text in c#
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1. COGood job. Excellent idea of support of optional blocks
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