Leetcode3---> Oldest string length with no repeating characters

title: given a string of strings, find the oldest string with no repeating characters in string.

Example:

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 asubstring, is "pwke" a subsequence and not a substring.

Ideas:

First, because we are looking for a non-repeating substring, we use the hashset structure to hold the characters that have been found, and the hashset holds all the characters from the pre to another position I in the middle.

Traverse the given string s from the beginning, each traversing one, to determine if there is a character in the hashset, assuming that it is traversed at the I position:

1: If not, the character is added to the HashSet, indicating that the length of the substring is increased by one, that is, the prev~i position of the string is no repeating character, update the length of the longest string max_str;

2: If there is, start looping through the prev position until the character that is equal to the I position character is found, and then prev points to the position of the character position +1, I continues the traversal.

Until I==len ends the traversal. However, the size of Max (I-prev, MAX_STR) should be computed at this time, the size of the MAX_STR is updated, and the final max_str is returned;

Extended:

Suppose the subject asks to find the oldest string without duplication, you need to use two variables to save the left and right position of the window, whenever the MAX_STR update, you need to update the window left and right position. Finally, use s.substring (left, right) to get the oldest string.

The code:

1  Public classSolution {2      Public intlengthoflongestsubstring (String s) {3         if(s = =NULL|| S.length () < 1)4             return0;5Hashset<character> set =NewHashset<character>();6         intPre = 0;//left edge of window7         intMax_str = Integer.min_value;//Longest string length8         inti = 0;//the right edge of the window9         intLen =s.length ();Ten          while(I <len) One         { A             if(Set.contains (S.charat (i)))//find the character that is equal to the I position, the pre points to the next position of the character, and restarts the window -             { -                 if(I-pre >max_str) theMax_str = i-Pre; -                  while(S.charat (PRE)! = S.charat (i))//until you find the character that is equal to the current character, then you can start the new round of window counting again -                 { - Set.remove (S.charat (pre)); +Pre + +; -                 } +Pre + +; A             } at             Else -             { - Set.add (S.charat (i)); -             } -i++; -         } inMax_str = Math.max (Max_str, I-pre);//I go backwards until the length of S is exceeded, and the size of the window is calculated at this time -         returnMax_str; to          +     } -}

Leetcode3---> Oldest string length with no repeating characters