[leedcode 131] Palindrome Partitioning

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

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

public class Solution {
    //DFS思想，getPart函数的意思是从s的start索引处开始求回文，DFS过程是，取start的i个字符，如果是回文，保存结果，从start+i开始遍历
    List<List<String>> res;
    List<String> seq;
    public List<List<String>> partition(String s) {
        res=new ArrayList<List<String>>();
        seq=new ArrayList<String>();
        getPart(s,0);
        return res;
    }
    public void getPart(String s,int start){
        if(start==s.length()){
            res.add(new ArrayList<String>(seq));
        }
        for(int i=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 boolean isPart(String s){
        int i=0;
        int j=s.length()-1;
        while(i<j){
            if(s.charAt(i)!=s.charAt(j))return false;
            i++;
            j--;
        }
        return true;
    }
    
    
}