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

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 
Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. 
             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

Java1

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int maxLength = 0, curStart = 0;
        Map<Character, Integer> lookup = new HashMap<>();
        for (int i = 0; i < s.length(); ++i) {
            char cur = s.charAt(i);
            if (lookup.containsKey(cur) && lookup.get(cur) >= curStart) {
                maxLength = Math.max(maxLength, i - curStart);
                curStart = lookup.get(cur) + 1;
            }
            lookup.put(cur, i);
        }
        maxLength = Math.max(maxLength, s.length() - curStart);
        return maxLength;
    }
}

Java2

效率更高一些，前提是但是假设char是8位

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int[] charRecentIndex = new int[256];
        for (int i = 0 ; i <charRecentIndex.length; ++i)
            charRecentIndex[i] = -1;
            
        int len = s.length(), curStart = 0, ret = 0;
        
        for (int i = 0 ; i < len; ++i){       
            char cur = s.charAt(i);
            if(curStart <= charRecentIndex[cur]){
                int preMaxLength = i - curStart;
                ret = Math.max(ret, preMaxLength);
                curStart = charRecentIndex[cur] + 1;
            }
            charRecentIndex[cur] = i;
        }
        ret = Math.max(ret, s.length() - curStart);
        return ret;
            
    }
}

Python

ord()返回字符的ASCII码

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        longest, start, visited = 0, 0, [False for _ in range(256)]
        for i, char in enumerate(s):
            if visited[ord(char)]:
                while char != s[start]:
                    visited[ord(s[start])] = False
                    start += 1
                start += 1
            else:
                visited[ord(char)] = True
            longest = max(longest, i - start + 1)
        return longest