131. Palindrome Partitioning

First, the topic

1, examining

　　

2. Analysis

Give a string, divide it into sub-strings, make its substring all palindrome, ask all the segmentation.

Second, the answer

1, Ideas:

Method One,

Recursive method is used for segmentation.

①, recursive, determine whether the current split substring is a palindrome, if, store the substring, and split the string, continue to recursion the remaining substrings;

②, recursive jump condition is: the character of the current string cut subscript i >= the length of the string;

     PublicList<list<string>>partition (String s) {List<List<String>> resultlist =NewArraylist<>(); BackTrack (Resultlist,NewArraylist<string> (), s, 0); returnresultlist; }    Private voidBackTrack (list<list<string>> resultlist, arraylist<string> curlist, String s,inti) {if(Curlist.size () > 0 && i >=s.length ()) {Resultlist.add (NewArraylist<string>(curlist)); return; }         for(intj = i; J < S.length (); J + +) {            if(Ispalindrome (S, I, J)) {Curlist.add (S.substring (i, J+ 1)); BackTrack (Resultlist, Curlist, S, J+ 1); Curlist.remove (Curlist.size ()+ W); }        }    }    Private BooleanIspalindrome (String S,intLintr) { while(L <r) {if(S.charat (l++)! = S.charat (r--))                return false; }        return true; }

Method Two,

Implemented with DP + DFS.

The use of a dynamic array of DP Records is a palindrome, in lieu of each recursive for palindrome judgment. DP[I][J] = true: Indicates that the string from subscript I to subscript j in S is a palindrome.

Then recursive for the DFS traversal, the s of all the Cut palindrome string records.

     PublicList<list<string>>Partition2 (String s) {List<List<String>> resultlist =NewArraylist<>(); Boolean[] DP =New Boolean[S.length ()][s.length ()];  for(inti = 0; I < s.length (); i++) {             for(intj = 0; J <= I; J + +) {                if(S.charat (j) = = S.charat (i) && (i-j <= 2 | | dp[j+1][i-1])) {Dp[j][i]=true; }}} partitionhelper (Resultlist,NewArraylist<string> (), DP, S, 0); returnresultlist; }        Private voidPartitionhelper (list<list<string>> resultlist, arraylist<string>ArrayList,Boolean[] DP, String s,inti) {if(Arraylist.size () > 0 && i >=s.length ()) {Resultlist.add (NewArraylist<>(arrayList)); return; }         for(intj = i; J < S.length (); J + +) {            if(Dp[i][j]) {Arraylist.add (S.substring (i, J+ 1)); Partitionhelper (Resultlist, ArrayList, DP, S, J+ 1); Arraylist.remove (Arraylist.size ()-1); }        }    }

131. Palindrome partitioning