leetcode3 Longest Substring Without Repeating Characters

"""
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.
"""
"""
解法一：双指针法，维护了一个滑动窗口
便于理解，实用性强
"""
class Solution1:
    def lengthOfLongestSubstring(self, s):
        res = 0
        l, r = 0, 0
        window = [] #双指针法，维护了一个滑动窗口
        while l < len(s) and r < len(s):
            if s[r] not in window:
                window.append(s[r])
                r += 1
                res = max(res, r - l)
            else:
                window.pop(0)
                l += 1
        return res
"""
解法二：用dict，与leetcode1结合类似
难点在于用一个下标不断更新dict中重复字符的值，
"""
class Solution2:
    def lengthOfLongestSubstring(self, s):
        d = dict()
        max_len = 0
        start = 0 #作为子串左端标记,一直向右
        for i in range(len(s)):
            if s[i] in d and d[s[i]] >= start:#！！！！
                start = d[s[i]] + 1 #!!!用start一直更新重复字符的新值
            temp = i - start + 1
            d[s[i]] = i
            max_len = max(max_len, temp)
        return max_len

"""
解法三：想用暴力法
Wrong Answer
Input
"abcabcbb"
Output
5
Expected
3
abcbb没符合这种情况
"""
class Solution3:
    def lengthOfLongestSubstring(self, s: str) -> int:
        max_len = 0
        for i in range(len(s)):
            for j in range(i + 1, len(s)):
                if s[i] == s[j]: #这逻辑有问题
                    j -= 1
                    break  # 很有用,找到第一个相等的跳出循环
            max_len = max(max_len, j - i + 1)
        return max_len