@aditya : I am wondering how many times 7 has occurred. Is it 1? Or is it 0?
Please take a moment before posting your solution, and think whether it is write or wrong! On Jul 6, 12:11 am, aditya kumar <[email protected]> wrote: > Q3. ans:70000000 i guess this is also a correct answer and no unique soln as > such > > On Wed, Jul 6, 2011 at 12:37 AM, aditya kumar > <[email protected]>wrote: > > > > > > > > > boolean palindromeCheck(String string) > > { > > len=string.length(); > > if((string.length()>1)) > > { > > if((string.charAt(0)==string.charAt(len-1))) > > { > > str=""; > > str=str+string.substring(1,(len-1)); > > palindromeCheck(str); > > } > > else > > { > > flag=false; > > } > > } > > return flag; > > } > > > THis also works fine if you dont want to use pointer > > > On Tue, Jul 5, 2011 at 9:37 PM, Azhar Hussain <[email protected]> wrote: > > >> For Q4: I think this is the optimal code > > >> int recurPalin(char *start, char *end) > >> { > >> if (end < start) > >> return true; > > >> if (*start != *end) > >> return false; > > >> return recurPalin(start+1, end-1); > >> } > > >> - > >> Azhar. > > >> On Tue, Jul 5, 2011 at 12:21 PM, vikas <[email protected]> wrote: > > >>> My program for Q4. > >>> // recursively find if a given string is palindrome > >>> bool IsPalindrome(string s, int start, int start2, bool flag) > >>> { > >>> bool flag1 = flag; > >>> if (start >= 0 && start2 < (s.Length)) > >>> { > >>> char c1 = s[start]; > >>> char c2 = s[start2]; > >>> if (c1.Equals(c2)) > >>> { > >>> if (start == 0 && start2 == s.Length - 1) { flag = > >>> true; } > >>> if (IsPalindrome(s, start - 1, start2 + 1, flag)) > >>> { > >>> flag1 = true; > >>> } > >>> } > >>> } > >>> return flag1; > >>> } > > >>> while calling > >>> if (s.Length % 2 != 0) > >>> { > >>> p.IsPalindrome(s, s.Length / 2 - 1, s.Length / 2 + 1, > >>> false); > >>> } > >>> else > >>> { > >>> p.IsPalindrome(s, s.Length / 2 - 1, s.Length / 2, false); > > >>> } > > >>> -- > >>> You received this message because you are subscribed to the Google Groups > >>> "Algorithm Geeks" group. > >>> To view this discussion on the web visit > >>>https://groups.google.com/d/msg/algogeeks/-/I6SVTB0o-uUJ. > > >>> To post to this group, send email to [email protected]. > >>> To unsubscribe from this group, send email to > >>> [email protected]. > >>> For more options, visit this group at > >>>http://groups.google.com/group/algogeeks?hl=en. > > >> -- > >> You received this message because you are subscribed to the Google Groups > >> "Algorithm Geeks" group. > >> To post to this group, send email to [email protected]. > >> To unsubscribe from this group, send email to > >> [email protected]. > >> For more options, visit this group at > >>http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
