131. Palindrome Partitioning

Simply happy, once through, time is not particularly good, I think it is ispalindrome use is the Stirngbuilder method is not the reason for the mathematical method

But very happy!!

1      PublicList<list<string>>partition (String s) {2list<list<string>> res =NewArraylist<list<string>>();3         if(s = =NULL|| S.length () = = 0) {4             returnRes;5         }6Helper (res, s, 0,NewArraylist<string>());7         returnRes;8     }9     Ten     Private voidHelper (list<list<string>> Res, String s,intStart, list<string>curstr) { One         intLen =s.length (); A         if(Start = =Len) { -Res.add (NewArraylist<string>(CURSTR)); -             return; the         } -          for(inti = 1; i + start <= len; i++) { -String sub = s.substring (Start, i +start); -             if(Ispalindrome (sub)) { + Curstr.add (sub); -Helper (res, S, start +I, curstr); +Curstr.remove (Curstr.size ()-1); A             } at         } -     } -      -     Private BooleanIspalindrome (String s) { -         if(s = =NULL|| S.length () = = 0) { -             return true; in         } -StringBuilder SB =NewStringBuilder (s); toString rev =sb.reverse (). toString (); +         returns.equals (rev.); -}

131. Palindrome Partitioning