LeetCode 93. Restore IP Addresses

原题链接在这里：https://leetcode.com/problems/restore-ip-addresses/

题目：

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)

题解：

IP规则, 一共有四段. 每段数字在[0,255], 但当一段起始是0时, 它唯一的合法规则就是单独一个0. 

DFS state: s, current start index, count item, count of segments in the item and res.

一共四段, dfs时用count计数现在是第几段. 到了第四段并且正好走到了s的末位, 把当前item加到res中.

Otherwise, for i in [1,3], get s.substring(start, start+i), of course start + i is within index range. Check if value of substring is legal. 

If yes, then continue DFS. 

Time Complexity: exponential.

Space: O(n). regardless res. 最多4层stack. each stack layer there is a String item. n = s.length().

AC Java:

class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<>();
        if(s == null || s.length() == 0){
            return res;
        }
        
        dfs(s, 0, 0, "", res);
        return res;
    }
    
    private void dfs(String s, int start, int count, String item, List<String> res){
        if(count > 4){
            return;
        }
        
        if(count == 4 && start == s.length()){
            res.add(item);
            return;
        }
        
        for(int i = 1; i < 4 && start + i <= s.length(); i++){
            String sub = s.substring(start, start + i);
            if(isLegal(sub)){
                dfs(s, start + i, count + 1, item + sub + (count == 3 ? "" : "."), res);
            }
        }
    }
    
    private boolean isLegal(String s){
        if(s.charAt(0) == '0'){
            return s.equals("0");
        }
        
        int val = Integer.valueOf(s);
        return val > 0 && val < 256;
    }
}

Time Complexity: exponential.

Space: O(n). regardless res.

AC Java:

class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<>();
        if(s == null || s.length() < 4){
            return res;
        }
        
        dfs(s, 0, new ArrayList<String>(), res);
        return res;
    }
    
    private void dfs(String s, int start, List<String> item, List<String> res){
        if(start == s.length() && item.size() == 4){
            res.add(String.join(".", item));
            return;
        }
        
        if(item.size() > 4){
            return;
        }
        
        for(int i = 1; i < 4 && start + i <= s.length(); i++){
            String can = s.substring(start, start + i);
            if(isLegal(can)){
                item.add(can);
                dfs(s, start + i, item, res);
                item.remove(item.size() - 1);
            }
        }
    }
    
    private boolean isLegal(String s){
        if(s.charAt(0) == '0'){
            return s.equals("0");
        }
        
        int val = Integer.valueOf(s);
        return val < 256 && val >= 0;
    }
}