93. 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)

题目含义：给定一个数字字符串，给出所有可能的IPV4地址

思路：[0，i),[i,j),[j,k),[k,s.length()-1)

由于s.substring（i，j）是包含位置i的字符而不包含位置j的字符，所以(1<=i<4),由于i的后面要至少保留两位给j和k，所以i<s.length()-2

j在i的后面1-3位，所以 i+1 <= j < i+4,由于j后面至少要保留1位给k，所以j<i<s.length()-1

k在j的后面1-3位，所以 j+1 <= k < j+4,由于k后面可以为空，所以保证数组不越界就可以了k<s.length()

 1     private boolean isValid(String num) {
 2         if (num.isEmpty() || num.length() > 3 || (num.charAt(0) == '0' && num.length() > 1) || Integer.parseInt(num) > 255)
 3             return false;
 4         return true;
 5     }
 6     
 7     public List<String> restoreIpAddresses(String s) {
 8         List<String> result = new ArrayList<>();
 9         for (int i = 1; i < 4 && i < s.length() - 2; i++) {
10             for (int j = i + 1; j < i + 4 && j < s.length() - 1; j++)
11                 for (int k = j + 1; k < j + 4 && k < s.length(); k++) {
12 
13                     String s0 = s.substring(0, i);
14                     String s1 = s.substring(i, j);
15                     String s2 = s.substring(j, k);
16                     String s3 = s.substring(k, s.length());
17                     if (isValid(s0) && isValid(s1) && isValid(s2) && isValid(s3)) {
18                         result.add(s0 + "." + s1 + "." + s2 + "." + s3);
19                     }
20                 }
21         }
22         return result;        
23     }