leetcode--Restore IP Addresses

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given "25525511135",

return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

This problem can be solved by bfs method

public class Solution {
    public ArrayList<String> restoreIpAddresses(String s) {
        ArrayList<String> result = new ArrayList<String>();
        int len = s.length();
        if(len >= 4 && len <= 12){
          result = bfs(s);        
        }
        return result;    
    }

    public ArrayList<String> bfs(String s){
        ArrayList<String> result = new ArrayList<String>();
        Queue<PairStrings> tempResult = new LinkedList<PairStrings>();
        tempResult.add(new PairStrings("", s));
        for(int i = 0; i < 4; ++i){
            Queue<PairStrings> temp = new LinkedList<PairStrings>();
            while(tempResult.peek() != null){
                PairStrings aPair = tempResult.poll();
                String aString = aPair.s2;
                int len = aString.length();
                if(len > 0){
                    for(int j = 1; j < 4 && j < len + 1; ++j){
                        String subs = aString.substring(0,j);
                        if((len - j <= 3 * ( 3 - i)) && checkValidation(subs)){
                            PairStrings newPair = new PairStrings(aPair.s1 + "." + subs, aString.substring(j, len));
                            temp.add(newPair);
                        }    
                    }
                }
            }
            tempResult = temp;
        }
        while(tempResult.peek() != null){
            PairStrings finalPairs = tempResult.poll();
            String finalString = finalPairs.s1;
            if(finalString.charAt(0) == '.')
                result.add(finalString.substring(1, finalString.length()));
        }
        return result;
    }

    public boolean checkValidation(String s){
        boolean canAppend = false;
        if(s.length() != 0){
            if(s.charAt(0) == '0')
                canAppend = (s.equals("0"));
            else{
                int num = Integer.parseInt(s);
                canAppend = (num >= 0 && num <= 255);
            }
        }
        return canAppend;
    }
}

class PairStrings{
    String s1;
    String s2;
    PairStrings(String s1, String s2){
        this.s1 = s1;
        this.s2 = s2;
    }
}

　　

 rewrite the above code in a simple way

public class Solution {
    public List<String> restoreIpAddresses(String s) {
		int len = s.length();
		List<String> validIP = new ArrayList<String>();
		if(len < 4 || len > 12)
			return validIP;
		dfs(s, "", validIP, 0);
		return validIP;
	}
	
	private void dfs(String s, String startIp, List<String> validIp, int rounds) {
		if(rounds == 3 && isValid(s)) // the last round
			validIp.add(startIp + s);
		else {
			for(int i = 1; i < 4 && i < s.length(); ++i) {
				String current = s.substring(0, i);
				if(isValid(current))
					dfs(s.substring(i), startIp + current+".", validIp, rounds + 1);
			}
		}
	}
	
	private boolean isValid(String s){
		if(s.charAt(0) == '0')
			return s.length() == 1;
		return s.length() <= 3 && Integer.parseInt(s) <= 255; 
	}
}