我一直在最長的子字符串上花費數(shù)小時而不重復(fù)字符 - LeetCode不重復(fù)字符的最長子串中等的給定一個字符串，找出最長的不重復(fù)字符的子串的長度。示例 1：Input: "abcabcbb"Output: 3 Explanation: The answer is "abc", with the length of 3. 示例 2：Input: "bbbbb"Output: 1Explanation: The answer is "b", with the length of 1.示例 3：Input: "pwwkew"Output: 3Explanation: 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.可以使用兩個指針混合 kadane 的算法來操作子數(shù)組來解決該問題class Solution:    def lengthOfLongestSubstring(self, s: str) -> int:        logging.debug(f"{list(enumerate(s))}")        slo = fas = 0  #slow as the fisrt character in a subarray which not duplicate, fast as the fast.                                  #relation: length = fas - slo        current = set()        glo = loc = 0        while fas < len(s):            logging.debug(f"pre_current: {current}, slow: {slo}, fast: {fas}")            if s[fas] not in current:                 current.add(s[fas]                loc = fas - slo                glo = max(glo, loc)                 fas +=1            else:                current.remove(s[slo])                slo += 1            logging.debug(f"post_current: {current}, slow: {slo}, fast: {fas} \n")        return glo測試用例    def test_g(self):        s = "abccefg"        answer = 4        check = self.solution.lengthOfLongestSubstring(s)        self.assertEqual(answer, check)解決方案很清楚慢速和快速交替移動$ python 3.LongestSubstring.py MyCase.test_gDEBUG [(0, 'a'), (1, 'b'), (2, 'c'), (3, 'c'), (4, 'e'), (5, 'f'), (6, 'g')]DEBUG pre_current: set(), slow: 0, fast: 0DEBUG post_current: {'a'}, slow: 0, fast: 1 DEBUG pre_current: {'a'}, slow: 0, fast: 1DEBUG post_current: {'b', 'a'}, slow: 0, fast: 2 DEBUG pre_current: {'b', 'a'}, slow: 0, fast: 2DEBUG post_current: {'b', 'c', 'a'}, slow: 0, fast: 3 ----------------------------------------------------------------------Ran 1 test in 0.001s作為結(jié)論，該解決方案采用了兩個指針技術(shù)和 Kadane 算法的思想。我假設(shè)作為初學者花費數(shù)小時調(diào)試后，最終有可能解決它。