3. 无重复字符的最长子串

"abc"，所以其

"b"

"wke"

for (int i = 0, j = 0; i < n; ++i) {
    while (j < i && check(j, i)) {
        ++j;
    }
    // 具体问题的逻辑
}

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        ss = set()
        i = ans = 0
        for j, c in enumerate(s):
            while c in ss:
                ss.remove(s[i])
                i += 1
            ss.add(c)
            ans = max(ans, j - i + 1)
        return ans

class Solution {
    public int lengthOfLongestSubstring(String s) {
        Set<Character> ss = new HashSet<>();
        int i = 0, ans = 0;
        for (int j = 0; j < s.length(); ++j) {
            char c = s.charAt(j);
            while (ss.contains(c)) {
                ss.remove(s.charAt(i++));
            }
            ss.add(c);
            ans = Math.max(ans, j - i + 1);
        }
        return ans;
    }
}

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        unordered_set<char> ss;
        int i = 0, ans = 0;
        for (int j = 0; j < s.size(); ++j) {
            while (ss.count(s[j])) ss.erase(s[i++]);
            ss.insert(s[j]);
            ans = max(ans, j - i + 1);
        }
        return ans;
    }
};

func lengthOfLongestSubstring(s string) int {
	ss := map[byte]bool{}
	i, ans := 0, 0
	for j := 0; j < len(s); j++ {
		for ss[s[j]] {
			ss[s[i]] = false
			i++
		}
		ss[s[j]] = true
		ans = max(ans, j-i+1)
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function (s) {
    const ss = new Set();
    let i = 0;
    let ans = 0;
    for (let j = 0; j < s.length; ++j) {
        while (ss.has(s[j])) {
            ss.delete(s[i++]);
        }
        ss.add(s[j]);
        ans = Math.max(ans, j - i + 1);
    }
    return ans;
};

public class Solution {
    public int LengthOfLongestSubstring(string s) {
        var ss = new HashSet<char>();
        int i = 0, ans = 0;
        for (int j = 0; j < s.Length; ++j)
        {
            while (ss.Contains(s[j]))
            {
                ss.Remove(s[i++]);
            }
            ss.Add(s[j]);
            ans = Math.Max(ans, j - i + 1);
        }
        return ans;
    }
}

function lengthOfLongestSubstring(s: string): number {
    const ss = new Set();
    let i = 0;
    let ans = 0;
    for (let j = 0; j < s.length; ++j) {
        while (ss.has(s[j])) {
            ss.delete(s[i++]);
        }
        ss.add(s[j]);
        ans = Math.max(ans, j - i + 1);
    }
    return ans;
}

class Solution {
    func lengthOfLongestSubstring(_ s: String) -> Int {
        var map = [Character: Int]()
        var currentStartingIndex = 0
        var i = 0
        var maxLength = 0
        for char in s {
            if map[char] != nil {
                if map[char]! >= currentStartingIndex {
                    maxLength = max(maxLength, i - currentStartingIndex)
                    currentStartingIndex = map[char]! + 1
                }
            }
            map[char] = i
            i += 1
        }
        return max(maxLength, i - currentStartingIndex)
    }
}

proc lengthOfLongestSubstring(s: string): int =
  var
    i = 0
    j = 0
    res = 0
    literals: set[char] = {}

  while i < s.len:
    while s[i] in literals:
      if s[j] in literals:
        excl(literals, s[j])
      j += 1
    literals.incl(s[i]) # Uniform Function Call Syntax f(x) = x.f
    res = max(res, i - j + 1)
    i += 1

  result = res # result has the default return value

use std::collections::HashSet;

impl Solution {
    pub fn length_of_longest_substring(s: String) -> i32 {
        let s = s.as_bytes();
        let mut set = HashSet::new();
        let mut i = 0;
        s.iter()
            .map(|c| {
                while set.contains(&c) {
                    set.remove(&s[i]);
                    i += 1;
                }
                set.insert(c);
                set.len()
            })
            .max()
            .unwrap_or(0) as i32
    }
}