Restore IP Address

题目：

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)

这个题，我觉得主要考察的是细节吧。因为我就漏了一个test case。 我错在：

Input:	"010010"
Output:	["0.1.0.010","0.1.00.10","0.1.001.0","0.10.0.10","0.10.01.0","0.100.1.0","01.0.0.10","01.0.01.0","01.00.1.0","010.0.1.0"]
Expected:	["0.10.0.10","0.100.1.0"]

思路上直接想到DFS。就用DFS解了。

    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<String>();
        if (s == null || s.length() == 0) {
            return res;
        }
        helper(0, 1, "", s, res);
        return res;
    }
    
    private void helper(int index, int seg, String temp, String s, List<String> res) {
        if (index >= s.length()) {
            return;
        }
        if (seg == 4) {
            if (isValid(s.substring(index))) {
                temp += s.substring(index);
                res.add(temp);
            }else{
                return;
            }
        }
        for (int i = 1; i < 4; i++) {
                if (index + i < s.length() && isValid(s.substring(index, index + i))) {
                    helper(index + i, seg + 1, temp + s.substring(index, index + i) + ".", s, res);
                }else{
                    return;
                }
        }
    }
    
    private static boolean isValid(String s) {
        if (s.length() <= 0 || s.length() > 3) {
            return false;
        }
        if (s.charAt(0) == '0' && s.length() != 1) {
            return false;
        }
        int n = Integer.parseInt(s);
        if (n < 0 || n > 255) {
            return false;
        }
        return true;
    }