LeetCode 93 Restore IP Addresses

Description

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)

Solution

1. 问题是要把一段字符串转换成可能的ip地址
2. 定义返回变量solutions
3. 传入restoreIp函数中

• restoreIp 函数有四个参数原始的ip字符串、 返回结果solutions、idx（当前ip的索引）、restored、count（用来记录切割的ip段， ip总共四段）
• count > 4 就返回
• count == 4 && idx == ip的长度，意味着完成了ip地址的转换，并将存储的变量restored 加入到 solutions中
• 进行for循环 i = 1 2 3

如果idx+i > ip.length() 跳出循环
s = ip.substring(idx, idx+i)
s以0开始而且长度大于1 || i = 3&&s>256 跳过下一条语句
递归调用restoreIp(ip, solutions, idx移动i, restored + s + （count == 3 ? "" : "."）, count +1);

Code

public class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> solutions = new ArrayList<>();
        restoreIp(s, solutions, 0, "", 0);
        return solutions;
    }
    
    private void restoreIp(String ip, List<String> solutions, int idx, String restored, int count) {
        if(count > 4) return;
        if(count == 4 && idx == ip.length()) {
            solutions.add(restored);
        }
        for(int i = 1; i < 4; i++) {
            if(idx + i > ip.length()) break;
            String s = ip.substring(idx, idx+i);
            if ((s.startsWith("0") && s.length() > 1) || (i == 3 && Integer.parseInt(s) >= 256)) 
                continue;
            
            restoreIp(ip, solutions, idx+i, restored+s+(count == 3 ? "" : "."), count+1);
        }
    }
}