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)

Solution1:

betters

  public List<String> restoreIpAddresses(String s) {    List<String> result = new ArrayList<>();    int len = s.length();    for (int i = 0; i < 4 && i < len - 2; i++) {      for (int j = i + 1; j < i + 4 && j < len - 1; 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)) {            result.add(s1 + "." + s2 + "." + s3 + "." + s4);          }        }      }    }    return result;  }  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;  }

Solution2:

  public List<String> restoreIpAddresses(String s) {    List<String> result = new ArrayList<>();    restoreIpAddresses(s, 0, result, "", 4);    return result;  }  private void restoreIpAddresses(String ip, int begin, List<String> result, String temp, int count) {    if (count == 0) {      if (begin == ip.length())        result.add(temp);      return;    }    for (int i = 1; i < 4; i++) {      if (begin + i > ip.length())        break;      String s = ip.substring(begin, begin + i);      if ((s.charAt(0) == '0' && s.length() > 1) || Integer.parseInt(s) > 255)        continue;      restoreIpAddresses(ip, begin + i, result, temp + s + (count == 1 ? "" : "."), count - 1);    }  }