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)

第一种解法，backtracking, 找出第一部分合法的IP, 剩余部分变成相似子问题。

public class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<String>();
        if(s.length() < 4 || s.length() > 12) return res;
        dfs(s, "", res, 1);
        return res;
    }
    
    public void dfs(String s, String temp, List<String> res, int count){
        if(count == 4 && isValid(s)){
            res.add(temp + s);
            return;
        }
        
        for(int i = 1; i < Math.min(4, s.length()); i++){
            String cur = s.substring(0, i);
            if(isValid(cur)){
                dfs(s.substring(i),  temp + cur + ".", res, count+1);
            }
        }
    }
    
    public boolean isValid(String s){
        if(s.charAt(0) == '0') return s.equals("0");
        int num = Integer.parseInt(s);
        return 0 < num && num < 256;
    }
}

2 这里IP的特性是最大数字不能超过255。比上个方法好的地方在于a+b+c+d= s.length()才会判断数字是否合法，避免了很多1+1+1+N这种不需要检查的情况。

public class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> ret = new ArrayList<>();
        
        StringBuffer ip = new StringBuffer();
        for(int a = 1 ; a < 4 ; ++ a)
        for(int b = 1 ; b < 4 ; ++ b)
        for(int c = 1 ; c < 4 ; ++ c)
        for(int d = 1 ; d < 4 ; ++ d)
        {
            if(a + b + c + d == s.length() )
            {
                int n1 = Integer.parseInt(s.substring(0, a));
                int n2 = Integer.parseInt(s.substring(a, a+b));
                int n3 = Integer.parseInt(s.substring(a+b, a+b+c));
                int n4 = Integer.parseInt(s.substring(a+b+c));
                if(n1 <= 255 && n2 <= 255 && n3 <= 255 && n4 <= 255)
                {
                    ip.append(n1).append('.').append(n2)
                        .append('.').append(n3).append('.').append(n4);
                    if(ip.length() == s.length() + 3) ret.add(ip.toString());
                    ip.delete(0, ip.length());
                }
            }
        }
        return ret;
    }
}