Leetcode 3. Restore IP Addresses

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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)

解法一：

1、IP地址的区域是从0到255

2、建立四个字符串并分别验证它们是否合理

public class Solution {    public List<String> restoreIpAddresses(String s) {        List<String> res = new ArrayList<>();        int len = s.length();        for(int i = 1; i < 4 ; i++){            for(int j = i+1; j < i+4; j++){                for(int k = j+1; k < j+4 && k < len; k++){                    String s1 = s.substring(0,i);                    String s2 = s.substring(i,j);                    String s3 = s.substring(j,k);                    String s4 = s.substring(k,len);                    if(isValid(s1) && isValid(s2) && isValid(s3) && isValid(s4)){                        res.add(s1+"."+s2+"."+s3+"."+s4);                    }                }            }        }        return res;    }    private boolean isValid(String s) {        if(s.length()>3 || s.length()==0 || (s.charAt(0)=='0' && s.length()>1) || Integer.parseInt(s)>255)            return false;        return true;    }}

解法二：

public class Solution {    public List<String> restoreIpAddresses(String s) {        List<String> res = new ArrayList<String>();        dfs(s, res, 0, "", 0);        return res;    }        private void dfs(String ip, List<String> res, int len, String temp, int count) {        if (count > 4) return;        if (count == 4 && len == ip.length()) res.add(temp);        for (int i = 1; i < 4; i++) {            if (len + i > ip.length()) break;            String s = ip.substring(len, len + i);            if ((s.startsWith("0") && s.length()>1) || (i==3 && Integer.parseInt(s) >= 256)) continue;            dfs(ip, res, len+i, temp + s + (count == 3 ? "" : "."), count + 1);        }    }

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