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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

最长不重复子串

Brute Force

有一个判断当前字符串为不重复的函数 boolean allUnique(String substring) .

然后两重循环求解，时间复杂度$O(n^3)$

sliding window

在暴力搜索中， boolean allUnique(String substring) 对于字串是从头开始搜索的。如果使用set的结构，可以将复杂度降到$O(n^2)$

Sliding Window Optimized

暴力搜索会产生很多不必要的操作，比如$s_{i,j}​$代表字符串 $i​$ 到 $j-1​$ 没有重复字串。则我们只需要判断第 $j​$ 个是否含于 $s_{i, j}​$ 即可。如果不包含，则 $s_{i,j}​$ 变为 $s_{i,j+1}​$ 。如果包含，则从包含的下标的下一位置开始(记录对应的位置)。

如下图所示：

下一坐标起始点即为2

这样就把时间复杂度降到了$O(n)$

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def lengthOfLongestSubstring(self, s):
  """
  :type s: str
  :rtype: int
  """
  used = {}
  max_length = start = 0
  for i, c in enumerate(s):
    if c in used and start <= used[c]:
      start = used[c] + 1
    else:   
      max_length = max(max_length, i - start + 1)
          
    used[c] = i
  
  return max_length