README_EN.md

3. Longest Substring Without Repeating Characters

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Description

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

 

Example 1:

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

Example 2:

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

Example 3:

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

Example 4:

Input: s = ""
Output: 0

 

Constraints:

• 0 <= s.length <= 5 * 104
• s consists of English letters, digits, symbols and spaces.

Solutions

Python3

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        i = j = res = 0
        chars = set()
        while i < len(s):
            while s[i] in chars:
                if s[j] in chars:
                    chars.remove(s[j])
                j += 1
            chars.add(s[i])
            res = max(res, i - j + 1)
            i += 1
        return res

Java

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

JavaScript

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

TypeScript

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;
  };

Swift

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)
    }
}

Go

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
}

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