LeetCode - 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)
Note: according to the feedback from the test cases, any part that is in the form of '0x' or '0xx' is not valid.
Recursion: Using DFS
from http://blog.csdn.net/u011095253/article/details/9158449
0-255的数字，转换成字符，即每个Part 可能由一个字符组成，二个字符组成，或者是三个字符组成。那这又成为组合问题了，dfs便呼之欲出
所以我们写一个For循环每一层从1个字符开始取一直到3个字符，再加一个isValid的函数来验证取的字符是否是合法数字，如果是合法的数字，我们再进行下一层递归，否则跳过。
Maybe we can convert String to list first.
https://discuss.leetcode.com/topic/24184/backtracking-solution-in-java-easy-to-understand/2
public List<String> restoreIpAddresses(String s) {
    List<String> res = new ArrayList<>();
    if(s == null || s.length() == 0 || s.length() > 12) {
        return res;
    }
    helper(s, res, new StringBuilder(), 0, 0);
    return res;
}

private void helper(String s, List<String> res, StringBuilder sb, int pos, int count) {
    if(pos == s.length() && count == 3) {
        res.add(sb.toString());
        return;
    }
    if(count == 3 || pos == s.length()) {
        return;
    }
    for(int i = pos; i < s.length(); i++) {
        String t = s.substring(pos, i+1);
        if(t.length() > 3 || t.length() > 1 && t.charAt(0) == '0' || Integer.valueOf(t) > 255) {
            break;
        }
        int len = sb.length();
        
        sb.append(t);
        if(i+1 != s.length()) {
            sb.append(".");
            helper(s, res, sb, i+1, count+1);
        } else {
            helper(s, res, sb, i+1, count);
        }
        sb.setLength(len);
    }
}
https://discuss.leetcode.com/topic/34366/java-recursive-backtracking-easy-to-read
public List<String> restoreIpAddresses(String s) {
    List<String> ret = new LinkedList<>();
    int[] path = new int[4];
    helper(ret, s, 0,  path, 0);
    return ret;
}

void helper(List<String> acc, String s, int idx, int[] path,  int segment){
    if(segment == 4 && idx == s.length() ){
        acc.add(path[0] + "." + path[1] + "."+ path[2] + "." + path[3]);
        return ;
    }else if(segment == 4 || idx == s.length() ){
        return ;
    }
    
    for(int len = 1; len <= 3 && idx + len <= s.length() ; len ++){
        int val = Integer.parseInt(s.substring(idx, idx + len));// not good
        // range check, no leading 0.
        if(val > 255 || len >= 2  && s.charAt(idx) == '0') 
            break; 
            
        path[segment] = val;
        helper(acc, s, idx + len, path, segment + 1);
        path[segment] = -1; // for debug. 
    }
}
https://discuss.leetcode.com/topic/41384/simple-java-solution-beating-100-of-java-submissions
private static final char DOT = '.';

public List<String> restoreIpAddresses(String s) {
    List<String> result = new ArrayList<String>();
    
    char[] digits = s.toCharArray();
    int len = s.length();
    char[] currIpAddr = new char[len+3];
    int pos = 0;
    generateIpAddresses(digits, 4, 0, len, currIpAddr, pos, result);
    
    return result;
}

private void generateIpAddresses(char[] digits, int remSegs, int start, int len, 
                                 char[] currIpAddr, int pos, List<String> result) {
        if(start == len && remSegs == 0) {
            result.add(String.valueOf(currIpAddr));
            return;
        }   
    
   //1. Checks for length of s too small
   //2. Maximum Length of the remaining segments. Since a sgemnt can be upto 3 digits
   // Length can not exceed 3x the remaining segments.
   //3. Minimum Length of s. Each segment has to be atleast 1 digit
    if((start > len) || ((len - start) > (3 * remSegs)) || ((len - start) < remSegs))
        return;
    
    if(remSegs < 4)
        currIpAddr[pos++] = DOT;
    
    int num = 0;
    
    for(int i = 0; i < Math.min(len-start, 3);i++) {
        num = (10*num) + (int)(digits[start+i] - '0');//\\
        
        if(i > 0 && num < 10)// leading 0 cases i = 1, then the number should be > 10.
            return;
        
        ////"010010"
        //Valid: ["0.10.0.10","0.100.1.0"]
        //Invalid: ["0.1.0.010","0.1.00.10","0.1.001.0","0.10.0.10","0.10.01.0","0.100.1.0",
        //"01.0.0.10","01.0.01.0","01.00.1.0","010.0.1.0"]
        
        if(num <= 255) {
            currIpAddr[pos+i] = digits[start+i];
            generateIpAddresses(digits, remSegs-1, start+i+1, len, currIpAddr, pos+i+1, result);
        }
    }
  }

http://www.shuatiblog.com/blog/2014/05/24/Restore-IP-Addresses/
public ArrayList<String> restoreIpAddresses(String s) {
    ArrayList<String> res = new ArrayList<String>();  
    if (s.length()<4||s.length()>12) return res;  
    dfs(s,"",res,0);  
    return res;  
}  

public void dfs(String s, String tmp, ArrayList<String> res, int count){  
    if (count == 3 && isValid(s)) {  
        res.add(tmp + s);  
        return;  
    }  
    for(int i=1; i<4 && i<s.length(); i++){  
        String substr = s.substring(0,i);  
        if (isValid(substr)){  
            dfs(s.substring(i), tmp + substr + '.', res, count+1);  
        }  
    }  
}  

public boolean isValid(String s){  
    if (s.charAt(0)=='0') return s.equals("0");  
    int num = Integer.parseInt(s);  
    return num<=255 && num>0;  
}
https://discuss.leetcode.com/topic/4742/very-simple-dfs-solution

http://www.jyuan92.com/blog/leetcode-restore-ip-addresses/
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public List<String> restoreIpAddresses(String s) {
    List<String> res = new LinkedList<String>();
    if (null == s || s.length() < 4 || s.length() > 12) {
        return res;
    }
    helper(res, s, new LinkedList<String>(), 0);
    return res;
}
private void helper(List<String> res, String s, List<String> list, int count) {
    if (list.size() == 4) {
        if (count == s.length()) {
            StringBuilder sb = new StringBuilder();
            for (String ip : list) {
                sb.append(ip);
                sb.append(".");
            }
            sb.deleteCharAt(sb.length() - 1);
            res.add(sb.toString());
        }
        return;
    }
    for (int i = count; i <= count + 3 && i < s.length(); i++) {
        String temp = s.substring(count, i + 1);
        if (isValid(temp)) {
            list.add(temp);
            helper(res, s, list, i + 1);
            list.remove(list.size() - 1);
        }
    }
}

  1. Here, we should realise that there is no need to do Backtracking after recursive, because every time when you do operation on String, it will create a new String rather than the original one.
  2. after recurive the first three part, we can just using the remain string to check for the fourth part
public List<String> restoreIpAddresses2(String s) {
    List<String> res = new LinkedList<String>();
    if (null == s || s.length() < 4 || s.length() > 12) {
        return res;
    }
    helper2(res, s, "", 0);
    return res;
}
private void helper2(List<String> res, String s, String str, int count) {
    if (count == 3 && isValid(s)) {
        res.add(str + s);
        return;
    }
    for (int i = 0; i < s.length() - 1 && i <= 3; i++) {
        String temp = s.substring(0, i + 1);
        if (isValid(temp)) {
            helper2(res, s.substring(i + 1), str + temp + ".", count + 1);
        }
    }
}
http://www.jiuzhang.com/solutions/restore-ip-addresses/
    public ArrayList<String> restoreIpAddresses(String s) {
        ArrayList<String> result = new ArrayList<String>();
        ArrayList<String> list = new ArrayList<String>();
        
        if(s.length() <4 || s.length() > 12)
            return result;
        
        helper(result, list, s , 0);
        return result;
    }
    
    public void helper(ArrayList<String> result, ArrayList<String> list, String s, int start){
        if(list.size() == 4){
            if(start != s.length())
                return;
            
            StringBuffer sb = new StringBuffer();
            for(String tmp: list){
                sb.append(tmp);
                sb.append(".");
            }
            sb.deleteCharAt(sb.length()-1);
            result.add(sb.toString());
            return;
        }
        
        for(int i=start; i<s.length() && i<= start+3; i++){
            String tmp = s.substring(start, i+1);
            if(isvalid(tmp)){
                list.add(tmp);
                helper(result, list, s, i+1);
                list.remove(list.size()-1);
            }
        }
    }
    
    private boolean isvalid(String s){
        if(s.charAt(0) == '0')
            return s.equals("0"); // to eliminate cases like "00", "10"
        int digit = Integer.valueOf(s);
        return digit >= 0 && digit <= 255;
    }


http://www.programcreek.com/2014/06/leetcode-restore-ip-addresses-java/

public List<String> restoreIpAddresses(String s) {
    ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
    ArrayList<String> t = new ArrayList<String>();
    dfs(result, s, 0, t);
 
    ArrayList<String> finalResult = new ArrayList<String>();
 
    for(ArrayList<String> l: result){
        StringBuilder sb = new StringBuilder();
        for(String str: l){
            sb.append(str+".");
        }
        sb.setLength(sb.length() - 1);
        finalResult.add(sb.toString());
    }
 
    return finalResult;
}
 
public void dfs(ArrayList<ArrayList<String>> result, String s, int start, ArrayList<String> t){
    //if already get 4 numbers, but s is not consumed, return
    if(t.size()>=4 && start!=s.length()) 
        return;
 
    //make sure t's size + remaining string's length >=4
    if((t.size()+s.length()-start+1)<4) 
        return;
 
    //t's size is 4 and no remaining part that is not consumed.
    if(t.size()==4 && start==s.length()){
        ArrayList<String> temp = new ArrayList<String>(t);
        result.add(temp);
        return;
    }
 
    for(int i=1; i<=3; i++){
        //make sure the index is within the boundary
        if(start+i>s.length()) 
            break;
 
        String sub = s.substring(start, start+i);
        //handle case like 001. i.e., if length > 1 and first char is 0, ignore the case.
        if(i>1 && s.charAt(start)=='0'){
            break;    
        }
 
        //make sure each number <= 255
        if(Integer.valueOf(sub)>255)
            break;
 
        t.add(sub);
        dfs(result, s, start+i, t);
        t.remove(t.size()-1);
    }
http://bangbingsyb.blogspot.com/2014/11/leetcode-restore-ip-addresses.html
于以s[i]开头的数字有3种可能：
1. s[i]
2. s[i : i+1]，s[i] !=0时
3. s[i : i+2]，s[i] != 0，且s[i : i+2] <= 255
    bool isValidNum(string s) {
        if(s.empty() || s.size()>3) return false;
        if(s[0]=='0' && s.size()!=1) return false;
        if(s.size()==3 && stoi(s)>255) return false;
        return true;
    }
https://leetcode.com/discuss/80435/java-recursive-backtracking-easy-to-read
public List<String> restoreIpAddresses(String s) { List<String> ret = new LinkedList<>(); int[] path = new int[4]; helper(ret, s, 0, path, 0); return ret; } void helper(List<String> acc, String s, int idx, int[] path, int segment){ if(segment == 4 && idx == s.length() ){ acc.add(path[0] + "." + path[1] + "."+ path[2] + "." + path[3]); return ; }else if(segment == 4 || idx == s.length() ){ return ; } for(int len = 1; len <= 3 && idx + len <= s.length() ; len ++){ int val = Integer.parseInt(s.substring(idx, idx + len)); // range check, no leading 0. if(val > 255 || len >= 2 && s.charAt(idx) == '0') break; path[segment] = val; helper(acc, s, idx + len, path, segment + 1); path[segment] = -1; // for debug. not needed } }
X. Iterative Version:
https://discuss.leetcode.com/topic/6304/my-concise-ac-java-code
the basic idea is to make three cuts into the string, separating it into four parts, each part contains 1~3 digits and it must be <255.
you don't need the fourth for loop with d. since d = len - (a+b+c) and
//java int D= Integer.valueof(s.substr(a+b+c)))
static List<String> restoreIpAddresses(String s) {
 List<String> ans = new ArrayList<String>();
 int len = s.length();
 for (int i = 1; i <=3; ++i){  // first cut
  if (len-i > 9) continue;    //  
  for (int j = i+1; j<=i+3; ++j){  //second cut
   if (len-j > 6) continue;   //    
   for (int k = j+1; k<=j+3 && k<len; ++k){  // third cut
    int a,b,c,d;                // the four int's seperated by "."
    a = Integer.parseInt(s.substring(0,i));  
    b = Integer.parseInt(s.substring(i,j)); // notice that "01" can be parsed into 1. Need to deal with that later.
    c = Integer.parseInt(s.substring(j,k));
    d = Integer.parseInt(s.substring(k));
    if (a>255 || b>255 || c>255 || d>255) continue; 
    String ip = a+"."+b+"."+c+"."+d;
    if (ip.length()<len+3) continue;  // this is to reject those int's parsed from "01" or "00"-like substrings
    ans.add(ip);
   }
  }
 }
 return ans;
}

https://discuss.leetcode.com/topic/3919/my-code-in-java
3-loop divides the string s into 4 substring: s1, s2, s3, s4. Check if each substring is valid. In isValid, strings whose length greater than 3 or equals to 0 is not valid; or if the string's length is longer than 1 and the first letter is '0' then it's invalid; or the string whose integer representation greater than 255 is invalid.
-- not efficient as the one below. - if s1 is not valid, no need to continue
public List<String> restoreIpAddresses(String s) { List<String> res = new ArrayList<String>(); int len = s.length(); for(int i = 1; i<4 && i<len-2; i++){ for(int j = i+1; j<i+4 && j<len-1; j++){ for(int k = j+1; k<j+4 && k<len; k++){ String s1 = s.substring(0,i), s2 = s.substring(i,j), s3 = s.substring(j,k), s4 = s.substring(k,len); if(isValid(s1) && isValid(s2) && isValid(s3) && isValid(s4)){ res.add(s1+"."+s2+"."+s3+"."+s4); } } } } return res; } public boolean isValid(String s){ if(s.length()>3 || s.length()==0 || (s.charAt(0)=='0' && s.length()>1) || Integer.parseInt(s)>255) return false; return true; }

public ArrayList<String> restoreIpAddresses(String s) {
    ArrayList<String> res = new ArrayList<String>();  
    if (s.length() > 12 || s.length() < 4) return res;
    for (int i = 1; i < 4; i ++) {
        String first = s.substring(0, i);
        if (! isValid(first)) continue;
        for (int j = 1; i + j < s.length() && j < 4; j ++) {
            String second = s.substring(i, i + j);
            if (! isValid(second)) continue;
            for (int k = 1; i + j + k < s.length() && k < 4; k ++) {
                String third = s.substring(i + j, i + j + k);
                String fourth = s.substring(i + j + k);
                if (isValid(third) && isValid(fourth)) 
                    res.add(first + "." + second + "." + third + "." + fourth);
            }
        }
    }
    return res;
}
https://leetcode.com/discuss/19296/my-concise-ac-java-code
static List<String> restoreIpAddresses(String s) { List<String> ans = new ArrayList<String>(); int len = s.length(); for (int i = 1; i <=3; ++i){ // first cut if (len-i > 9) continue; for (int j = i+1; j<=i+3; ++j){ //second cut if (len-j > 6) continue; for (int k = j+1; k<=j+3 && k<len; ++k){ // third cut int a,b,c,d; // the four int's seperated by "." a = Integer.parseInt(s.substring(0,i)); b = Integer.parseInt(s.substring(i,j)); // notice that "01" can be parsed into 1. Need to deal with that later. c = Integer.parseInt(s.substring(j,k)); d = Integer.parseInt(s.substring(k)); if (a>255 || b>255 || c>255 || d>255) continue; String ip = a+"."+b+"."+c+"."+d; if (ip.length()<len+3) continue; // this is to reject those int's parsed from "01" or "00"-like substrings ans.add(ip); } } } return ans; }
http://www.darrensunny.me/leetcode-restore-ip-addresses/
Alternatively, we can consider each part iteratively by explicitly determining the positions of the dots.
public ArrayList<String> restoreIpAddresses(String s) {
        ArrayList<String> result = new ArrayList<String>();
        if (s == null)
            return result;
        int n = s.length();
        if (n < 4 || n > 12)
            return result;
        // For each possible first part
        for (int i = 1; i <= 3; i++) {      // At most three digits at each part
            if (n-i > 9)    // The number of digits for remaining three parts should not exceed 9
                continue;
            if (s.charAt(0) == '0' && i > 1)    // 0x or 0xx is not allowed
                break;
            String first = s.substring(0, i);   // String before the first dot
            if (Integer.parseInt(first) <= 255) {   // The first part is within proper range
                // For each possible second part given the first part
                for (int j = i+1; j <= i+3 && j < n; j++) {
                    if (n-j > 6)    // The number of digits for remaining two parts should not exceed 6
                        continue;
                    if (s.charAt(i) == '0' && j > i+1)
                        break;
                    String second = s.substring(i, j);  // String after the first dot and before the second
                    if (Integer.parseInt(second) <= 255) {  // The second part is within proper range
                        // For each possible third part given the second part
                        for (int k = j+1; k <= j+3 && k < n; k++) {
                            if (n-k > 3)    // The number of digits for remaining part should not exceed 3
                                continue;
                            if (s.charAt(j) == '0' && k > j+1)
                                break;
                            String third = s.substring(j, k);   // String after the second dot and before the third
                            if (Integer.parseInt(third) <= 255) {   // The third part is within proper range
                                // If the fourth part is a single digit, or it does not begin with a '0' and
                                // it is within proper range, we have got a valid IP address
                                if (k == n-1 || s.charAt(k) != '0' && Integer.parseInt(s.substring(k)) <= 255)
                                    result.add(first+'.'+second+'.'+third+'.'+s.substring(k));
                            }
                        }
                    }
                }
            }
        }
        return result;
    }
https://leetcode.com/discuss/88736/who-can-beat-this-code
vector<string> restoreIpAddresses(string s) { vector<string> ret; string ans; for (int a=1; a<=3; a++) for (int b=1; b<=3; b++) for (int c=1; c<=3; c++) for (int d=1; d<=3; d++) if (a+b+c+d == s.length()) { int A = stoi(s.substr(0, a)); int B = stoi(s.substr(a, b)); int C = stoi(s.substr(a+b, c)); int D = stoi(s.substr(a+b+c, d)); if (A<=255 && B<=255 && C<=255 && D<=255) if ( (ans=to_string(A)+"."+to_string(B)+"."+to_string(C)+"."+to_string(D)).length() == s.length()+3) ret.push_back(ans); } return ret; }
Also refer to http://yucoding.blogspot.com/2013/04/leetcode-question-83-restore-ip.html
Read full article from LeetCode - Restore IP Addresses | Darren's Blog
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