LeetCode – Longest Substring Without Repeating Characters (Java)

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

Analysis

The basic idea to solve this problem is using an extra data structure to track the unique characters in a sliding window. Both an array and a hash set work for this purpose.

Java Solution 1

The first solution is like the problem of "determine if a string has all unique characters" in CC 150. We can use a flag array to track the existing characters for the longest substring without repeating characters.

public int lengthOfLongestSubstring(String s) {
        if(s==null)
            return 0;
	boolean[] flag = new boolean[256];
 
	int result = 0;
	int start = 0;
	char[] arr = s.toCharArray();
 
	for (int i = 0; i < arr.length; i++) {
		char current = arr[i];
		if (flag[current]) {
			result = Math.max(result, i - start);
			// the loop update the new start point
			// and reset flag array
			// for example, abccab, when it comes to 2nd c,
			// it update start from 0 to 3, reset flag for a,b
			for (int k = start; k < i; k++) {
				if (arr[k] == current) {
					start = k + 1; 
					break;
				}
				flag[arr[k]] = false;
			}
		} else {
			flag[current] = true;
		}
	}
 
	result = Math.max(arr.length - start, result);
 
	return result;
}

Java Solution 2 - HashSet

Using a HashSet can simplify the code a lot.

/*
  pwwkew
i | 
j |
 
i |
j   |
 
i   |
j   |
 
*/
public int lengthOfLongestSubstring(String s) {
    if(s==null||s.length()==0){
        return 0;
    }
 
    HashSet<Character> set = new HashSet<>();
    int result = 1;
    int i=0; 
    for(int j=0; j<s.length(); j++){
        char c = s.charAt(j);
        if(!set.contains(c)){
            set.add(c);
            result = Math.max(result, set.size());
        }else{
            while(i<j){
                if(s.charAt(i)==c){
                    i++;
                    break;
                }
 
                set.remove(s.charAt(i));
                i++;
            }
        }    
    }
 
    return result;
}

Category >> Algorithms

If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:

<pre><code> 
String foo = "bar";
</code></pre>

Jesus Ledesma

public static String subStringMasLargo(String vector) { String subStringMasLargo = ""; String cadena = String.valueOf(vector.charAt(0));
for(int i=1; i subStringMasLargo.length()) { subStringMasLargo = cadena; cadena = String.valueOf(vector.charAt(i)); } else { cadena = String.valueOf(vector.charAt(i)); } } } return subStringMasLargo; } public static boolean laContiene(String cadena, String letra) { for (int i=0; i<cadena.length(); i++) { if(String.valueOf(cadena.charAt(i)).equals(letra)) { return true; } }
return false; }
Jesus Ledesma

public static String subStringMasLargo(String vector) { String subStringMasLargo = ""; String cadena = String.valueOf(vector.charAt(0));
for(int i=1; i subStringMasLargo.length()) { subStringMasLargo = cadena; cadena = String.valueOf(vector.charAt(i)); } else { cadena = String.valueOf(vector.charAt(i)); } } } return subStringMasLargo; } public static boolean laContiene(String cadena, String letra) { for (int i=0; i<cadena.length(); i++) { if(String.valueOf(cadena.charAt(i)).equals(letra)) { return true; } }
return false; }
Jesus Ledesma

public static String subStringMasLargo(String vector) { String subStringMasLargo = ""; String cadena = String.valueOf(vector.charAt(0));
for(int i=1; i subStringMasLargo.length()) { subStringMasLargo = cadena; cadena = String.valueOf(vector.charAt(i)); } else { cadena = String.valueOf(vector.charAt(i)); } } } return subStringMasLargo; } public static boolean laContiene(String cadena, String letra) { for (int i=0; i<cadena.length(); i++) { if(String.valueOf(cadena.charAt(i)).equals(letra)) { return true; } }
return false; }
Bukary Kandeh

Nice catch, Ash. I haven’t had time to look at your fix but I can visually see how that test case will fail. I’ll see if there is a fix with better efficiency when I get some time 🙂
Ashwini Singh

Hi Bukary,

This code will not work for “dvdf”

output should be 3 -> “vdf”
instead the output will be 2 ->”dv” or “df”
Bukary Kandeh

Java One-Loop Solution. See my short program that solves this problem in one loop and more efficient than the above solution. Time O(n)
public static int lengthOfLongestSubstring(String str){ Set set = new HashSet(); int longest=0, check=0;
for(int i=0; i < str.length(); i++){ if(!set.contains(str.charAt(i))){ set.add(str.charAt(i)); check++; }else { longest = Math.max(check,longest); set.clear(); check = 1; } } longest = Math.max(check,longest); return longest; }
abhishek pandey

In the Java Solution 2: What purpose does the start variable serve? I could not understand this part: while(start<i&&s.charAt(start)!=c){
set.remove(s.charAt(start));
start++;
}
start++;
Please help me
Sushanta Kumar Sahoo

for(j=i+1;jmax)—— showing error in this line in eclipse. could you tell me what is the error is about?
ryanlr

Time of this solution is N^3, isn’t it?
Anirudh Jayan

import java.util.*;
public class Subst {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println(“Enter String”);
String word=sc.next(),maxWord=””,newWord=””;
int l=word.length(),i,j,max=0;
for(i=0;i<l;i++){
newWord=word.substring(i);
for(j=i+1;jmax){
max=j-i;
maxWord=word.substring(i,j);
}
}
System.out.println(maxWord);
}
}
Gaurav Gupta

Below is a solution that is in O(n) time complexity and O(1) space.
This solution uses hash table of size 26 (number of alphabets).
I have tested it for as many cases as I could think of. Have a look:

#include
#include
main(){
char st[] = “geeksforgeeks”;
int hash[26] = {0};

int i, j = -1, max = 0, res = 1;

for (i = 0; i < strlen(st); i++) {
if ( (j != -1 && hash[st[i] – 'a'] < j) || (j == -1 && hash[st[i] – 'a'] == 0)) {
if (res < i – j) res = i – j;
} else {
j = hash[st[i] – 'a'];
}
hash[st[i] – 'a'] = i;
}
printf("%dn", res);
}
Ping Zhang

Here is my javascript solution with great details:

var lengthOfLongestSubstring = function(s) {
const valueIdxHash = {};
var longest = 0;
var startIdx = 0;
for(var i = 0; i < s.length; i++) {
if(s[i] in valueIdxHash) {
startIdx = Math.max(startIdx, valueIdxHash[s[i]] + 1);
}
valueIdxHash[s[i]] = i;
longest = Math.max(longest, i – startIdx + 1);
}
return longest;
};

Checkout my blog for detailed explanation: https://algorithm.pingzhang.io/String/longest_substring_without_repeating_characters.html
vuqar dadalov

This is My Solution

public class Solution {
public int lengthOfLongestSubstring(String s) {

if(s.trim().length() == 0) {
return 0;
}

char [] strArr = s.toCharArray();

HashMap map = new HashMap();
int maxLen = 0;
int firstIndex = 0;
int lastIndex = 0;

for(int i=0; i maxLen){
maxLen = i – firstIndex;
}

if(map.get(strArr[i]) >= firstIndex){
firstIndex = map.get(strArr[i]) + 1;
}

map.remove(map.get(strArr[i]));
}
map.put(strArr[i],i);
lastIndex = i;
}

if((lastIndex – firstIndex + 1) > maxLen){
maxLen = lastIndex – firstIndex + 1;
}

return maxLen;
}
}
Mel

static int lengthOfLongestSubstring(String s){ if(s==null || s.length()==0) return 0;
boolean []flag=new boolean[256]; int count=0; int max=Integer.MIN_VALUE; for(int i=0;imax){ max=count; } flag[c]=false; count=0; }
} return max; }
Badrinath

simple Solution is to add every character to Hash set and get the size of it. As Set doesn’t store duplicates this is the easiest and simple way to get the count.

below is the code
static int longest(String s){ int result = 0; int a = s.length(); Set set = new HashSet(); for(int i=0; i<a-1 ; i++){ set.add(s.charAt(i)); } System.out.println("Length of the string is" + set.size()); return result; }
Milan

//time : O(n) space : O(n)
public class Solution { public int lengthOfLongestSubstring(String s) { int start = 0,max = 0,i =0; Map lookup = new HashMap(); for (;i= start) { max=Math.max(max,i-start); start=lookup.get(ch)+1; } lookup.put(ch,i); } max=Math.max(max,i-start); return max; } }
jeevs

public class longestsubstring {

public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in=new Scanner(System.in);
ArrayList a=new ArrayList();
String s=in.next();
String temp=””;
String target=””;
for(int i=0;itarget.length())
target=temp;
a.clear();
temp=””;
}
}
System.out.print(target);
}
Eugene Arnatovich

Yep. That was my misunderstanding… Hope now it is better…
Larry Okeke

simple solution

import java.util.*; public class longest_substring{ /* Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1. */ public static void main(String[] args){ System.out.println("longest " +get_longest(args[0])); } public static int get_longest(String str){ HashSet wordSet = new HashSet(); int longest = 0; for(int i = 0; i < str.length(); i++){ for(int j = i; j longest){ longest = wordSet.size(); } wordSet.clear(); break; } } } return longest; } }
Larry Okeke

Lol! Are you kidding?
Eugene Arnatovich

Simplest solution:
//Longest Substring Without Repeating Characters (Java) public int solution(String s) { Set set = new LinkedHashSet(); for (char c : s.toCharArray()) { set.add(c); } return set.size(); }
Adrian

Following code is O(N) time complexity and constant space complexity O(1).

public int LongestSubstringWithoutRepeatingCharacters(String src) { if (src == null || src.Length == 0) return 0; // v[j] stores i position of the src[i] character, where j = src[i] int[] v = new int[256]; // Initialize visited state for (int i = 0; i < 256; ++i) v[i] = -1; v[src[0]] = 0; // start position of the longest string of non repeating characters int start = 0; // max length of string with non repeating characters so far int max_len = 1; for (int i = 1; i < src.Length; ++i) { if (v[src[i]] == -1) { v[src[i]] = i; ++max_len; } else { int cur_len = i - start; if (max_len < cur_len) max_len = cur_len; // move start of the substring with non repeating characters // after already visited(repeated) character start = v[src[i]] + 1; } }
return max_len; }
Truong Khanh Nguyen

Here is my O(N) time solution in Java http://www.capacode.com/string/longest-substring-without-repeating-characters/
Ankit Shah

Following works and it also prints the substring
CRH

Why is the time complexity mentioned as n^3 for Java Solution 2 above? It should be O(n). The first loop is O(n) and in that main loop i is getting pushed back to an already visited location. So effectively we may end up looping through the whole loop 2n times. That still means O(n), then why n^3 is mentioned.

Please help!
pm

public class LCSWIthoutRepeating{
public static void main(String args[]){
String str = “aaaaapritikamehta”;
LinkedHashMap hash = new LinkedHashMap();
int maxLen =0;
int max =0;
int num =0;
int arr[] = new int[26];
for(int i =0;i<str.length();i++){
char c = str.charAt(i);
if(hash.get(c) == null){
hash.put(c, i);
max++;
}else{
maxLen = i – hash.get(c)max?maxLen:max);
}

}
southie

O(n) solution with LinkedHashSet JAVA !

public class LongestPalindromeWithoutRepChar {

public static int lengthOfLongestSubstring(String s) {
int maxCount=0;
if(s.length() < 2) return s.length();
Set hs = new LinkedHashSet();

for(int i=0;i maxCount) maxCount = hs.size();
Iterator itr = hs.iterator();
Character curr = null;
while(c!=curr)
{
curr = itr.next();
itr.remove();
}
}
}

return maxCount;
}
}
Holden

Why time complexity of second approach is O(n^3)?
NB****

void LargestNonRepeatedSubStr() { string str = "abcabcbb"; int max_length = 0; int i = 0; while (i < str.length()) { set check_set; int j = i; int length = 0; while (j < str.length()) { if (check_set.find(str[j]) != check_set.end()) { break; } else { check_set.insert(str[j]); length++; } j++; } check_set.clear(); max_length = std::max(max_length, length); i++; } cout << "The max length is: " << max_length << "n"; }
Sahil Singla

It is required. Consider the case “dabcabcde”. If ‘map.get(arr[i]) < j' is not done then, for the iteration with "i=7" it would be computed that "d" has already contained in the HashMap and also the logic inside the "else" will fail and compute to give 'curr=7'.
dtReloaded

List list = new ArrayList();

for(int i=0 ; i<a.length();i++){

if(!list.contains(a.charAt(i))){

list.add(a.charAt(i));

}

else

break;

}

System.out.println("The String Lengtht" +a.length());
Satish

public static String uniqueCharSubstring(String str) {
String result = “”;
int len = str.length();
HashMap map = new HashMap();
char[] c = str.toCharArray();
int right = 0, max = 0;
for (int left = 0; left < len; left++) {
while (right max) {
max = right – left;
result = str.substring(left, right + 1);
}
right++;
}
}
return result;
}
Vimukthi Weerasiri

Thanks. Agree. Updated the code with some refactoring 🙂 Intention was to initialize it one prior to the rightPointer. Comments are welcome!
test

second = -1 is pretty weird.
Vimukthi Weerasiri

A simpler O(n) solution.

public int lengthOfLongestSubstring(String s) {
HashMap map = new HashMap();
if (s == null || s.length() == 0) return 0;
int first = 0, second = -1, answer = 0;
while (first != s.length()) {
Integer previousOccurrence = map.put(s.charAt(first), first);
if (previousOccurrence != null) second = Math.max(second, previousOccurrence);
answer = Math.max(answer, first – second);
first++;
}
return answer;
}
Edmond Hong

I don’t think you need add the statement: ‘map.get(arr[i]) < j'
Sahil Singla

I believe this is O(n):

public static int lengthOfLongestSubstring(String s) {
if(s==null)
return 0;
char[] arr = s.toCharArray();
int pre = 0;
HashMap map = new HashMap();
int j = 0;
int curr = 0;
for (int i = 0; i < arr.length; i++) {
if (!map.containsKey(arr[i]) || map.get(arr[i]) < j) {
map.put(arr[i], i);
curr++;
} else {
pre = Math.max(pre, curr);
j = map.get(arr[i]) + 1;
curr = i + 1 – j;
map.put(arr[i], i);
}
}
return Math.max(pre, curr);
}
Adil qureshi

can you explain why second one in On3 .. i cant get it … thanks
Taoufik Bdiri

For the solution using the HashMap, I think the following line :
i = map.get(arr[i]);
should be changed to :
i = map.get(arr[i])+1;
Otherwise you will get an infinite loop ?
Parag Chaudhari

Better solution: http://www.geeksforgeeks.org/length-of-the-longest-substring-without-repeating-characters/
Mehmet

What is the time and space complexity of first solution?
ryanlr

No, it doesn’t. For example, given “abac”, when it meets the second “a”, aux should not be set to “a”, instead it should be set to “ba”.
Tiago

I think the following works too:

static public String nonRepeated(String input){

char[] charAInput = input.toCharArray();
String aux = “”;
String resultSubStr = “”;

if(input.length() == 1)
return input;

for(int i =0; i < charAInput.length; i++ ){

if(!aux.contains("" + charAInput[i])){

aux = aux.concat("" + charAInput[i]);

}
else
aux = "" + charAInput[i];

if(resultSubStr.length() < aux.length())
resultSubStr = aux;
}

System.out.println(resultSubStr);

return resultSubStr;

}
ryanlr

You solution is wrong. maxStart is not updated correctly.
ryanlr

Your code is not right, for example, “aab”.
ryanlr

Thanks. I think the problem is fixed now.
ryanlr

Your code only compares the two consecutive characters, it can not handle case like “a”.
Arpit Pandey

“bbabceabcdbb” fails with this string returns a wrong string
Tarun Singh

I think Following is O(n) Solution 😀

public static int lengthOfLongestSubstring(String s) {

char[] arr = s.toCharArray();

int pre = 0;

int k=0;

boolean flag = true;

HashMap map = new HashMap();

for (int i = 0; i pre)

pre=l;

k = map.get(arr[i]);

map.put(arr[i],i);

k++;

}

}

if(flag) pre = arr.length;

return pre;

}
neckBeardThePirate

how is solution two O(n^3)?
Zaga

So I came up with an O(n) solution in python, please help me detect in what cases it can possibly fail

#put test string here

string = “eadsabcadcfef”

indexOfSubsequence = 0

tempIndexOfSubsequence = 0

lenghtOfSubsequence = 0

tempLenghtOfSubsequence = 0

tableOfCharsIndex = [-1] * 26

for i in xrange(0, len(string)):

indexOfChar = ord(string[i]) – 97

if tableOfCharsIndex[ indexOfChar ] == -1 or tableOfCharsIndex[ indexOfChar ] lenghtOfSubsequence :

lenghtOfSubsequence = tempLenghtOfSubsequence

indexOfSubsequence = tempIndexOfSubsequence

else:

if tempLenghtOfSubsequence > lenghtOfSubsequence :

lenghtOfSubsequence = tempLenghtOfSubsequence

indexOfSubsequence = tempIndexOfSubsequence

tempIndexOfSubsequence = i

tempLenghtOfSubsequence = 1

tableOfCharsIndex[ indexOfChar ] = i

print string[ indexOfSubsequence : indexOfSubsequence + lenghtOfSubsequence ]

print lenghtOfSubsequence
Zy

Hi Ryan,

I like your site a lot, but I noticed that the recently-added social area on every post is blocking the sight.

Hopefully this can be fixed and I can enjoy reading your site again.
Wally Osmond

public static int getNoRepeatLen (String str) {

int max=0,tLen=0;

for(int i=1;i<str.length();i++) {

if(str.charAt(i-1) == str.charAt(i)){

max=Math.max(tLen+1,max);

tLen=0;

} else {

tLen++;

}

}

return Math.max(max,tLen);

}
Wally Osmond

Hey guys,

Isn’t the best solution just a simple for loop with an if and some counters? Perhaps my solution is too simple and crashes or returns the results incorrectly somehow??? I tested it for the longest string and the beginning/end/middle and I believe it works.

public static int getNoRepeatLen (String str) {

int max=0,tLen=0;

for(int i=1;i<str.length();i++) {

if(str.charAt(i-1) == str.charAt(i)){

max=Math.max(tLen+1,max);

tLen=0;

} else {

tLen++;

}

}

return Math.max(max,tLen);

}
Jianwei Chen

this is simpler.
public static lls(String s) {
HashMap map = new HashMap();

int maxStart = 0;

int pre = 0;

for(int i = 0; i pre) {

pre = map.size();

maxStart = map.get(c);

}

}

map.put(c, i);

}

return s.substring(maxStart, maxStart+pre);
}
Krzysztof Rajda

Proposed solution could be improved. You don’t need to clear the whole map and change i.

private static String calculateFast(String text) {
Map map = new HashMap();

int maxStart = 0;
int maxEnd = 1;
int currentStart = 0;

for (int i = 0; i maxEnd – maxStart) {
maxStart = currentStart;
maxEnd = i;
}
currentStart = firstOccurrenceIndex + 1;
}
map.put(c, i);
}

if (text.length() – currentStart > maxEnd – maxStart) {
maxStart = currentStart;
maxEnd = text.length();
}

return text.substring(maxStart, maxEnd);
}
chitti

this code handles:

public String getLongestSubStringWithoutRepeatedChar (String str) {
{
List list = new ArrayList();
String longestSoFar =””;
int count = 0;
int longestSize=0;

//str.length()+1 for capturing longest string at the end.
for (int i = 0; i < str.length()+1; i++) {
if (i longestSize) {
longestSize=count;
longestSoFar = getLongestString(list);
System.out.println(longestSoFar);
}
count=0;
list.clear();

}
}

return longestSoFar;
}
}

private String getLongestString (List list) {
StringBuilder sb = new StringBuilder();
for (Character c : list) {
sb.append(c);
}
return sb.toString();
}
chitti

corrected one:

public String getLongestSubStringWithoutRepeatedChar (String str) {
{
List list = new ArrayList();
String longestSoFar =””;
int count = 0;
int longestSize=0;

for (int i = 0; i longestSize) {
longestSize=count;
longestSoFar = getLongestString(list);
System.out.println(longestSoFar);
}
count=0;
list.clear();

}
}
return longestSoFar;
}
}

private String getLongestString (List list) {
StringBuilder sb = new StringBuilder();
for (Character c : list) {
sb.append(c);
}
return sb.toString();
}
chitti

Does not work for string “eadsabcadcfef”
Alex

How about this O(n) solution? It’s easy to understand and allows for the retrieval of the longest substring as well:

public static int lengthOfLongestSubstring(String word) {

int start = 0;

int end = 0;

int maxStart = 0;

int maxEnd = 0;

int[] seen = new int[256];

for (int i = 0; i < 256; i++) {

seen[i] = -1;

}

for (int i = 0; i = start) {

start = seen[c] + 1;

}

end++;

seen[c] = i;

if (end – start > maxEnd – maxStart) {

maxEnd = end;

maxStart = start;

}

}

// Return word.substring(maxStart, maxEnd) for actual substring

return maxEnd – maxStart;

}
santhoshvai

I am sorry.. i misunderstood LIS.. this http://ideone.com/hcLYgy should fix it
Jonathan Young

abcadeftgh ==> bcadeftgh (this is the longest substring without repeating element, and its length is 9)
santhoshvai

I found the mistake, see this http://ideone.com/WMXFOw

the LIS for “abcadeftgh” is 7
Jonathan Young

This doesn’t work if you pass “abcadeftgh” to your method. It should output 9 instead of 6
santhoshvai

I have written a much simpler version than version 2 (python).. easier to understand. its here, http://ideone.com/BHJlvS
santhoshvai

whats the need for ” i = h.get(target[i]); ” ? It should work without it
sudhansu jena

Hi guys My solution simple and works fine.

public String getLongestSubStringWithoutRepeatedChar(String str){

ArrayList list=new ArrayList();

String longestSoFar=””;

int count=0;

for(int i=0;i<str.length();i++){

if(!list.contains(str.charAt(i))){

list.add(str.charAt(i));

}else{

if(count<list.size()){

count=list.size();

longestSoFar=getLongestString(list);

list.clear();

}

}

}

return longestSoFar;

}

private String getLongestString(ArrayList list) {

StringBuilder sb=new StringBuilder();

for(Character c:list){

sb.append(c);

}

return sb.toString();

}
Cheng Wang

My solution but no idea why it is not correct:

public class Solution {

public int lengthOfLongestSubstring(String s) {

// if the string is empty or has only one character

if (s.length() < 2) return s.length();

HashMap hashMap = new HashMap();

int j = 0;

int maxLen = 1;

for (int i = 0; i < s.length(); i++) {

if (!hashMap.containsKey(s.charAt(i))) {

hashMap.put(s.charAt(i), i);

} else {

maxLen = Math.max(maxLen, i – j);

j = hashMap.get(s.charAt(i)) + 1;

hashMap.put(s.charAt(i), i);

}

}

return Math.max(maxLen, s.length() – j);

}

}
Coder

Solution 2 cannot pass the OJ because of the Time limit.
amit malik

Updated Tia solution to

1. Find all patterns with no repeating characters.

2. Find the longest pattern and its size.

public static void printPatternWithNoReaptingCharsInString(String s) {

char[] arr = s.toCharArray();

Map charMap = new HashMap();

int size = 0;

List uniquePatterns = new ArrayList();

StringBuilder str = new StringBuilder();

for (int i = 0; i < arr.length; i++) {

if (!charMap.containsKey(arr[i])) {

charMap.put(arr[i], i);

} else {

if (size <= charMap.size()) {

size = charMap.size();

if (str.length() <= size) {

str.delete(0, str.length()).append(

s.substring(i – size, i));

uniquePatterns.add(str.toString());

}

}

i = charMap.get(arr[i]);

charMap.clear();

}

}

System.out.println("Longest Pattern with no repeating characters:"

+ str.toString() + " with size " + size);

System.out.println("There are in total " + uniquePatterns.size()

+ " patterns with no repeating characters.");

}

Below method call:

printUniquePatterninString("abcdaaabcdaabcdebbpqrstuvaaa");

will print:

Longest Pattern with no repeating characters:bpqrstuva with size 9

There are in total 6 patterns with no repeating characters.
play_boy

Tia’s solution is good.
I have the same solution but something different.
The time : O(n)
The space may be : O(n)
and it can output the maxlength substring. (you need to delete the comment tag, and run it)
like this:

public static int lengthOfLongestSubstring(String s) {
char[] arr = s.toCharArray();
int maxlength = 0;
int oldstart = 0;
int newstart = 0;
// int compare_times = 0; // compare times

HashMap num = new HashMap(); // recode char and the char index
HashMap map = new HashMap(); // recode index and the index of char

for (int i = 0; i < arr.length; i++) {
// compare_times++; //
if (num.get(arr[i]) != null) {
int end = num.get(arr[i]) + 1;
for (int j = newstart; j < end; j++)
{ // compare times will less than s.length()
num.remove(map.get(j));
map.remove(j);
}
newstart = end;
if (maxlength < num.size()) {
maxlength = num.size();
oldstart = newstart;
}
}
num.put(arr[i], i);
map.put(i, arr[i]);
}
if (maxlength < num.size()) {
maxlength = num.size();
oldstart = newstart;
}
// System.out.println("compare_times:" + compare_times); // output the compare times
// System.out.println(s.substring(oldstart, oldstart + maxlength)); // output the longest substring
return maxlength;
}
neal

this solution has time limited issue in leetcode.
ryanlr

good solution, thanks.
ryanlr

It’s the same.
lee

string longestStr = string.Empty;

int length = str.Length;

int i = 0;

Loop:

string tempLongestStr = string.Empty;

for (int j = i; j longestStr.Length)

{

longestStr = tempLongestStr;

}

i = i + tempLongestStr.IndexOf(temp) + 1;

goto Loop;

}

else

{

tempLongestStr += temp;

if (tempLongestStr.Length > longestStr.Length)

{

longestStr = tempLongestStr;

}
}

}

Console.WriteLine(longestStr);
Console.ReadLine();
junmin

public static int LongestSubStrWithUniqueChars(String s){
if(s==null || s.isEmpty())
return 0;

char[] chars = s.toCharArray();

int current = 0;//current pos in string
int length = 0; //current longest substr length
int pos = 0; //current longest substr starting pos

for(current =0 ; current<s.length(); current++){
pos = current;
boolean[] found = new boolean[256];
while(current<s.length() && !found[chars[current]]){
found[chars[current]] = true;
current++;
}
length = Math.max(length, current – pos);
current = pos;
}
return length;
}
tia

Here is my solution with hash:

public int solution(String s) {
// IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. char[] target = s.toCharArray(); int pre = 0; Hashtable h = new Hashtable(); for(int i =0; i h.size() ? pre : h.size(); i = h.get(target[i]);
h.clear(); } } return Math.max(pre, h.size()); }

LeetCode – Longest Substring Without Repeating Characters (Java)

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