[Leedcode 131] Palindrome Partitioning

Given A string s, partition s such that every substring of the partition are a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab" ,
Return

  [    ["AA", "B"],    ["A", "a", "B"]  ]

 Public classSolution {//Dfs idea, the Getpart function means starting at the start index of S to find a palindrome, the DFS process is, take the start of the I character, if it is a palindrome, save the results, from the start+i began to traverseList<list<string>>Res; List<String>seq;  PublicList<list<string>>partition (String s) {res=NewArraylist<list<string>>(); Seq=NewArraylist<string>(); GetPart (s),0); returnRes; }     Public voidGetPart (String S,intstart) {        if(start==s.length ()) {Res.add (NewArraylist<string>(seq)); }         for(intI=start+1;i<=s.length (); i++){            if(Ispart (s.substring (start,i))) {Seq.add (s.substring (start,i));                GetPart (S,i); Seq.remove (Seq.size ()-1); }        }    }     Public BooleanIspart (String s) {intI=0; intJ=s.length ()-1;  while(i<j) {            if(S.charat (i)!=s.charat (j))return false; I++; J--; }        return true; }        }

[Leedcode 131] Palindrome Partitioning