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:
        i = ans = 0
        chars = set()
        for j, c in enumerate(s):
            while c in chars:
                chars.remove(s[i])
                i += 1
            chars.add(c)
            ans = max(ans, j - i + 1)
        return ans

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

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

function lengthOfLongestSubstring(s: string): number {
    // 滑动窗口+哈希表
    let left = -1;
    let maxLen = 0;
    let hashTable = new Map();
    for (let right = 0; right < s.length; right++) {
        let cur = s.charAt(right);
        if (hashTable.has(cur)) {
            left = Math.max(left, hashTable.get(cur));
        }
        hashTable.set(cur, right);
        maxLen = Math.max(maxLen, right - left);
    }
    return maxLen;
}

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)
    }
}

func lengthOfLongestSubstring(s string) int {
	window := make(map[byte]int)
	n := len(s)
	ans := 0
	left, right := 0, 0
	for right < n {
		b := s[right]
		right++
		window[b]++
		for window[b] > 1 {
			window[s[left]]--
			left++
		}
		ans = max(ans, right-left)
	}
	return ans
}

func max(x, y int) int {
	if x > y {
		return x
	}
	return y
}

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

    }
};

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
    }
}