共计 1145 个字符，预计需要花费 3 分钟才能阅读完成。
Given a string containing only digits, restore it by returning all possible valid IP address combinations.
Example:
Input: “25525511135”
Output: [“255.255.11.135”, “255.255.111.35”]
难度：medium
题目：给定一字符串仅包含数字，恢复该字符串为所有可能合法的 IP 地址组合。
思路：递归
Runtime: 2 ms, faster than 91.11% of Java online submissions for Restore IP Addresses.Memory Usage: 34.8 MB, less than 0.90% of Java online submissions for Restore IP Addresses.
class Solution {
public List<String> restoreIpAddresses(String s) {
List<String> result = new ArrayList();
if (null == s || s.length() < 4 || s.length() > 12) {
return result;
}
restoreIpAddresses(s, 0, 0, “”, result);
return result;
}
private void restoreIpAddresses(String s, int i, int cnt, String str, List<String> result) {
int sLength = s.length();
if (i >= sLength && cnt >= 4) {
result.add(str.substring(0, str.length() – 1));
return;
}
int i1 = (i + 1) <= sLength ? Integer.parseInt(s.substring(i, i + 1)) : -1;
if (i1 >= 0 && cnt < 4) {
restoreIpAddresses(s, i + 1, cnt + 1, str + i1 + “.”, result);
}
int i2 = (i + 2) <= sLength ? Integer.parseInt(s.substring(i, i + 2)) : -1;
if (i2 >= 10 && i2 <= 99 && cnt < 4) {
restoreIpAddresses(s, i + 2, cnt + 1, str + i2 + “.”, result);
}
int i3 = (i + 3) <= sLength ? Integer.parseInt(s.substring(i, i + 3)) : -1;
if (i3 >= 100 && i3 <= 255 && cnt < 4) {
restoreIpAddresses(s, i + 3, cnt + 1, str + i3 + “.”, result);
}
}
}
