One brush 146 force buckle hot question-3 longest substring without repeated characters (m)

 subject ：
 Given a string  s , Please find out that there are no duplicate characters in it   Longest substrings   The length of .
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 Example ：
 Input : s = "abcabcbb"
 Output : 3 
 explain :  Because the longest substring without repeating characters is  "abc", So its length is  3.
 Example  2:

 Input : s = "bbbbb"
 Output : 1
 explain :  Because the longest substring without repeating characters is  "b", So its length is  1.
 Example  3:

 Input : s = "pwwkew"
 Output : 3
 explain :  Because the longest substring without repeating characters is  "wke", So its length is  3.
      Please note that , Your answer must be   Substring   The length of ,"pwke"  Is a subsequence , Not substring .
 Example  4:

 Input : s = ""
 Output : 0
 
 Tips ：

0 <= s.length <= 5 * 104
s  By the English letters 、 Numbers 、 Symbols and spaces 
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 reflection ：
 label ： The sliding window 
 The time complexity of violent solution is high , Will reach  O(n^2)O(n 2), Therefore, the sliding window method is adopted to reduce the time complexity 

 Define a  map  Data structure storage  (k, v), among  key  The value is the character ,value  Value is character position  +1,
 Add  1  Indicates that the character is not repeated until after the character position 

 We define the starting position of non repeating substring as  start, End position is  end
 With  end  Go back and forth , Will meet with  [start, end]  If the characters in the interval are the same , In this case, the character is used as  key  value ,
 Get its  value  value , And update the  start, here  [start, end]  There are no repeating characters in the interval 
 Whether updated or not  start, Will update it  map  Data structure and results  ans.
 Time complexity ：O(n)O(n)
 Code 
Java
class Solution {
    
    public int lengthOfLongestSubstring(String s) {
    
        if (s == null || s.length() == 0) return 0;
        int res = 0;
        Map<Character, Integer> map = new HashMap<>();
        for (int start = 0, end = 0; end < s.length(); end++) {
    
            char ch = s.charAt(end);
            if (map.containsKey(ch)) {
    
                start = Math.max(map.get(ch), start);
            }
            map.put(ch, end + 1);
            res = Math.max(res, end - start + 1);
        }
        return res;
    }
}

LC