No.3 longest Substring without repeating characters

Longest Substring without repeating characters

• Total accepted:167158
• Total submissions:735821
• Difficulty:medium

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb" , the answer "abc" is, which the length is 3.

Given "bbbbb" , the answer "b" is, with the length of 1.

Given "pwwkew" , the answer "wke" is, with the length of 3. Note that the answer must was a substring, is "pwke" a subsequence and not a substring.

Ideas for solving problems from: http://www.geeksforgeeks.org/length-of-the-longest-substring-without-repeating-characters/

1  Public classNum3 {2     /*3 * Method One: Brute force search, Complexity O (n^3)4      */5      Public intlengthoflongestsubstring (String s) {6         if(s = =NULL|| S.length () = = 0){7             return0 ;8         }9 String Sub;Ten          for(intSublen = S.length (); Sublen > 0; sublen--){ One              for(intStartIndex = 0; StartIndex <= (s.length ()-sublen); startindex++){ A                 //lists all substrings, and then determines if the substring satisfies a duplicate -                 if(StartIndex! = (S.length ()-Sublen)) { -Sub = s.substring (StartIndex, startindex+Sublen); the}Else{ -Sub =s.substring (startIndex); -                 } -                 if(!Isrepeat (sub)) { +                     returnSublen; -                 } +             } A         } at          -         return1 ; -     } -      -     Private BooleanIsrepeat (String s) { -          for(inti = 1; I < s.length (); i++){ in             if(S.substring (i). Contains (S.substring (i-1, i))) { -                 return true ; to             }  +         } -         return false ; the     } *      $     /*Panax Notoginseng * Method Two: Using hash method plus dynamic programming solution -      */ the      Public intlengthOfLongestSubstring2 (String s) { +         if(s = =NULL|| S.length () = = 0){ A             return0 ; the         } +         intCur_len = 1;//lenght of current substring -         intMax_len = 1 ; $         intPrev_index;//Previous Index $         int[] visited =New int[256] ; -         Char[] arr =S.tochararray (); -         /*Initialize the visited array as-1, 1 is used to the indicate that character have not been visited yet.*/ -          for(inti = 0; I < 256; i++){WuyiVisited[i] = 1 ; the         } -         /*Mark first character as visited by storing the index Wu of first character in visited array.*/ -Visited[arr[0]] = 0 ; About          $         /*Start from the second character. First character is - already processed (Cur_len and Max_len are initialized - as 1, and Visited[arr[0]] is set*/ -          for(inti = 1; i < arr.length; i++){ APrev_index =Visited[arr[i]]; +              the             /*If The current character isn't present in the - already processed substring or it is not part of $ the current NRCS and then do cur_len++*/ the             if(Prev_index = =-1 | | I-cur_len >Prev_index) { thecur_len++ ; the}Else{ the                 /*Also, when we are changing the NRCS, we - should also check whether length of the in previous NRCS was greater than Max_len or the Not .*/ the                 if(Cur_len >Max_len) { AboutMax_len =Cur_len; the                 } the                 //Update the index of current character theCur_len = i-Prev_index;  +             } -              theVisited[arr[i]] =i;Bayi         } the          the         //Compare The length of last NRCS with Max_len and -         //update Max_len if needed -         if(Cur_len >Max_len) { theMax_len =Cur_len; the         } the          the         returnMax_len; -          the     } the  the}

No.3 longest Substring without repeating characters