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)

Have you been asked this question in an interview? 

难度为3的题，属于排列的题目。
有几点细节容易出错：
1 递归的将数字串分成四个部分，每个部分满足0<=p<=255。要注意一些边界case，比如010是没有意思的，“0.10.010.1”. 
 最好用一个专门的方法来判断，比较清晰。
2 

if (!(value.charAt(0) != '0' && Integer.parseInt(value) <= 255)) {

把上面这个写成了下面的样子，一直没看出来哪里出错

if (!(value.charAt(0) != 0 && Integer.parseInt(value) <= 255)) {

注意char 和 int 的0 是不一样的
3 由于IPaddress  里需要加 . ,最好单独用一个arraylist 来处理，把输出单独拿出来写，可以让思路清晰，也不容易出错 

public class Solution {
    public List<String> restoreIpAddresses(String s) {
        if (s == null) {
            return null;
        }
        List<String> result = new ArrayList<String>();
        ArrayList<String> address = new ArrayList<String>();
        restore(result, address, s, 0);
        return result;
    }
    private void restore(List<String> res, ArrayList<String> address, String s, int startPosition) {
        if (address.size() == 4 && startPosition >= s.length()) {
            res.add(printAddress(address));
            return;
        } else if (address.size() == 4 || startPosition >= s.length()) {
            return;
        }
        for (int i = 1; i <= 3; i++) {
            if (startPosition + i <= s.length()) { 
                String value = s.substring(startPosition, startPosition + i);
                if (i > 1) {
                    if (!(value.charAt(0) != '0' && Integer.parseInt(value) <= 255)) {
                        break;
                    }
                }
                address.add(value);
                restore(res, address, s, startPosition + i);
                address.remove(address.size() - 1);
            }
        }
    }
    private String printAddress(ArrayList<String> address) {
        StringBuilder sb = new StringBuilder();
        sb.append(address.get(0));
        for (int i = 1; i < address.size(); i++) {
            sb.append('.');
            sb.append(address.get(i));
        }
        return sb.toString();
    }
}

别人解法：

public class Solution {
    public ArrayList<String> restoreIpAddresses(String s) {
        ArrayList<String> result = new ArrayList<String>();
        ArrayList<String> list = new ArrayList<String>();
        
        if(s.length() <4 || s.length() > 12)
            return result;
        
        helper(result, list, s , 0);
        return result;
    }
    
    public void helper(ArrayList<String> result, ArrayList<String> list, String s, int start){
        if(list.size() == 4){
            if(start != s.length())
                return;
            
            StringBuffer sb = new StringBuffer();
            for(String tmp: list){
                sb.append(tmp);
                sb.append(".");
            }
            sb.deleteCharAt(sb.length()-1);
            result.add(sb.toString());
            return;
        }
        
        for(int i=start; i<s.length() && i<= start+3; i++){
            String tmp = s.substring(start, i+1);
            if(isvalid(tmp)){
                list.add(tmp);
                helper(result, list, s, i+1);
                list.remove(list.size()-1);
            }
        }
    }
    
    private boolean isvalid(String s){
        if(s.charAt(0) == '0')
            return s.equals("0"); // to eliminate cases like "00", "10"
        int digit = Integer.valueOf(s);
        return digit >= 0 && digit <= 255;
    }
}