首页 > itarticle > [leetcode] 3. longest substring without repeating characters

[leetcode] 3. longest substring without repeating characters

admin 11月 14, 2020 0

meidum

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

Example 1:

1
2
3

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

Example 2:

1
2
3

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.

class :
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        
        if not s:
            return 0
        memo = set()
        self.max_length = float('-inf')
        for i in range(len(s)):
            if self.max_length >= len(s) - i:
                break
            self.helper("", i, s, memo)
        return self.max_length
    
    def helper(self, path, ind, string, memo):
        self.max_length = max(self.max_length, len(path))
        if ind >= len(string):
            return

        if string[ind] not in path:
            path+=string[ind]
            # print(path)
            self.helper(path, ind+1, string, memo)
        return
    
class Solution2:
    # sliding window
    # O(2n) time O(min(m, n)) space. m is 26, n is len of str
    def lengthOfLongestSubstring(self, s):
        if not s:
            return 0
        i = j = 0
        res = 0
        memo = set()
        while i < len(s) and j < len(s):
            if s[j] not in memo:
                memo.add(s[j])
                res = max(res, j - i + 1)
                j+=1
            else:
                memo.remove(s[i])
                i += 1
        return res
    
class Solution:
    # sliding window optimized using map
    # O(n) time O(min(m, n)) space. m is 26, n is len of str
    def lengthOfLongestSubstring(self, s):
        if not s:
            return 0
        res = 0
        i = j = 0
        dic = {}
        while j < len(s):
            if s[j] in dic:
                i = max(dic[s[j]], i)
            res = max(res, j - i + 1)
            dic[s[j]] = j + 1
            j+=1
        return res