leetcode3 Longest Substring Without Repeating Characters

 1 """
 2 Given a string, find the length of the longest substring without repeating characters.
 3 Example 1:
 4 Input: "abcabcbb"
 5 Output: 3
 6 Explanation: The answer is "abc", with the length of 3.
 7 Example 2:
 8 Input: "bbbbb"
 9 Output: 1
10 Explanation: The answer is "b", with the length of 1.
11 Example 3:
12 Input: "pwwkew"
13 Output: 3
14 Explanation: The answer is "wke", with the length of 3.
15              Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
16 """
17 """
18 解法一：双指针法，维护了一个滑动窗口
19 便于理解，实用性强
20 """
21 class Solution1:
22     def lengthOfLongestSubstring(self, s):
23         res = 0
24         l, r = 0, 0
25         window = [] #双指针法，维护了一个滑动窗口
26         while l < len(s) and r < len(s):
27             if s[r] not in window:
28                 window.append(s[r])
29                 r += 1
30                 res = max(res, r - l)
31             else:
32                 window.pop(0)
33                 l += 1
34         return res
35 """
36 解法二：用dict，与leetcode1结合类似
37 难点在于用一个下标不断更新dict中重复字符的值，
38 """
39 class Solution2:
40     def lengthOfLongestSubstring(self, s):
41         d = dict()
42         max_len = 0
43         start = 0 #作为子串左端标记,一直向右
44         for i in range(len(s)):
45             if s[i] in d and d[s[i]] >= start:#！！！！
46                 start = d[s[i]] + 1 #!!!用start一直更新重复字符的新值
47             temp = i - start + 1
48             d[s[i]] = i
49             max_len = max(max_len, temp)
50         return max_len
51 
52 """
53 解法三：想用暴力法
54 Wrong Answer
55 Input
56 "abcabcbb"
57 Output
58 5
59 Expected
60 3
61 abcbb没符合这种情况
62 """
63 class Solution3:
64     def lengthOfLongestSubstring(self, s: str) -> int:
65         max_len = 0
66         for i in range(len(s)):
67             for j in range(i + 1, len(s)):
68                 if s[i] == s[j]: #这逻辑有问题
69                     j -= 1
70                     break  # 很有用,找到第一个相等的跳出循环
71             max_len = max(max_len, j - i + 1)
72         return max_len