Recursive, DFS difficult to understand ah feel good

It said recursive encoding is an easy to understand, but the feeling is not so easy.

Today brush leetCode encountered a problem division palindrome string, as follows:

Given a string s, s is divided into a number of sub-strings, each sub-string is a string palindrome.

S return all the possible partitioning scheme.

Example:

输入: "aab"
输出:
[
["aa","b"],
["a","a","b"]
]

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/palindrome-partitioning
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

Among them the use of DFS and backtracking solution

The complete code is as follows

 

package com.LeeCode;

import java.util.ArrayList;
import java.util.List;

public class VariesPalindrome {
        List<List<String>>list=new ArrayList<List<String>>();
        String s;
        public List<List<String>> partition(String s) {
            //从头到尾递归+回溯。
            this.s=s;
            //这个是满足的每一个集合
            List<String>ll=new ArrayList<String>();
            dfs(ll,0);
            return list;
        }
        public void dfs(List<String>ll,int index){
            if(index==s.length())
            {
                list.add(new ArrayList<String>(ll));
                return;
            }
            // i is because a single character is palindromic substring from index 
            for (int I = index; I <s.length (); I ++) 
            { 
                // If palindromic 
                IF (isPalindrome (index, I)) { 

                    // the current palindromic substring s (index, i) added to the list 
                    ll.add (s.substring (index, I +. 1)); 
                    DFS (LL, I +. 1); 
                    // add the palindromic substring into place. The above and added into the same palindrome is a palindrome substring substring. 
                    ll.remove (ll.size () -. 1); 
                } 

            } 
        } 
        public Boolean isPalindrome (int Start, End int) { 
            the while (Start <End) { 
                ! IF (s.charAt (Start) = s.charAt (End) ) 
                    return false; 
                Start ++;
                end--;
            }
            return true;
        }

    public static void main(String[] args) {
        VariesPalindrome variesPalindrome=new VariesPalindrome();
        String jj="abnnss";
        System.out.println(variesPalindrome.partition(jj));
    }
}