[lintcode] - 426 Restore IP Addresses

这道题意思大概就是给一串数字，把它所有合法的IP地址列出来，并且不care排列顺序。事实上这道题有个坑，如果你的答案的顺序与testcase的不一致照样报错。

结题思路应该很容易想到，递归回溯就行了，每次选择当前掩码的数字位数，位数不合法的忽略，合法的跳到下一段掩码继续递归即可，递归最深处保留结果。注意判断递归停止条件。

/**
* @author: decaywood
* @date: 2016/1/27 9:58
*
* Given a string containing only digits,
* restore it by returning all possible valid IP address combinations.
*
* Example
* Given "25525511135", return
*
* [
*  "255.255.11.135",
*  "255.255.111.35"
* ]
* Order does not matter.
*
*/
public class RestoreIPAddresses {

public ArrayList<String> restoreIpAddresses(String s) {
ArrayList<String> arrayList = new ArrayList<>();
doRestore(arrayList, 0, 0, s, "");
return arrayList;
}

private void doRestore(ArrayList<String> addresses, int index, int level, String str, String candidate) {
if(level > 4) return;
level++;
if(index == str.length() && level == 5) {
addresses.add(candidate.substring(0, candidate.length() - 1));
return;
}
for (int i = 1; i <= 3; i++) {
if(index + i > str.length()) continue;
String sub = str.substring(index, index + i);
if(sub.length() > 1 && sub.charAt(0) == '0') continue;
int num = Integer.parseInt(sub);
if (num <= 255) {
doRestore(addresses, index + i, level, str, candidate + num + ".");
}
}
}

public static void main(String[] args) {
new RestoreIPAddresses().restoreIpAddresses("010010");
}

}