G-Face via prepare:chucked palindrome

Given a string, find out the maximum number of chunked palindrome, the normal palindrome is ABCCBA, chunked palindrome is defined as: For example, Volvo, can be the VO division, (VO) (l) (VO), Then it's a palindrome. Returns the maximum number of chunk to be implemented. For example, AAAAAA can be (AAA) (AAA), but the maximum chunk quantity should be (a) (a) (a) (a) (a) (a) to 6

This is a greedy problem, starting with the sides of the string, using I and J to record where the current scan is, using Prev_i and Prev_j to record the next character of the position of chunk I and J. Finally sweep to the middle to determine whether there is no extra chunk.

Time complexity O (n^2) with O (N) complexity in the inner layer string.equals

1  PackageChunkedpalindrome;2 3  Public classSolution {4      Public intcountchunk (String str) {5         if(str==NULL|| Str.length () ==0)return0;6         intsum = 0;7         intL=0, R=str.length ()-1;8         intPreL = L, Prer =R;9          while(L <r) {TenString left = str.substring (PreL, l+1); OneString right = Str.substring (R, prer+1); A             if(Left.equals (right)) { -PreL = l+1; -Prer = R-1; theSum + = 2; -             } -l++; -r--; +         } -         if(PreL <= prer) sum+=1; +         returnsum; A     } at      -  -     /** -      * @paramargs -      */ -      Public Static voidMain (string[] args) { in         //TODO auto-generated Method Stub -Solution Sol =Newsolution (); to         intres = Sol.countchunk ("aaaaaa"); + System.out.println (res); -     } the  *}

G-Face via prepare:chucked palindrome