LeetCode 3无重复最长子串：滑动窗口实现超时问题解析

我在解决LeetCode的《无重复字符的最长子串》题目，题目要求：给定字符串s，找出其中无重复字符的最长子串的长度。约束条件为：0 ≤ s.length ≤ 5*10^4，s包含英文字母、数字、符号和空格。

我使用滑动窗口算法实现的代码如下：

def lengthOfLongestSubstring(str):
    # define base case
    if (len(str) < 2): return len(str)

    # define pointers and frequency counter
    left = 0
    right = 0
    freqCounter = {} # used to store the character count
    maxLen = 0

    while (right < len(str)):
        # adds the character count into the frequency counter dictionary
        if (str[right] not in freqCounter):
            freqCounter[str[right]] = 1
        else:
            freqCounter[str[right]] += 1
            # print (freqCounter)
        # runs the while loop if we have a key-value with value greater than 1. 
        # this means that there are repeated characters in the substring. 
        # we want to move the left pointer by 1 until that value decreases to 1 again. E.g., {'a':2,'b':1,'c':1} to {'a':1,'b':1,'c':1}
        while (len(freqCounter) != right-left+1):
        # while (freqCounter[str[right]] > 1): ## Time Limit Exceeded Error
            print(len(freqCounter), freqCounter)
            freqCounter[str[left]] -= 1
            # remove the key-value if value is 0
            if (freqCounter[str[left]] == 0):
                del freqCounter[str[left]]
            left += 1
        
        maxLen = max(maxLen, right-left+1)
        # print(freqCounter, maxLen)
        right += 1
    
    return maxLen

print(lengthOfLongestSubstring("abcabcbb")) # 3 'abc'

我遇到了一个困惑：当使用while (freqCounter[str[right]] > 1):作为内层循环的判断条件时，提交代码会出现Time Limit Exceeded超时错误；但换成while (len(freqCounter) != right-left+1):就能正常通过测试。我原以为前者是O(1)的字典元素访问操作，效率应该更高，却没想到反而慢很多。同时我也想知道，当前的滑动窗口实现是否还有优化的空间。

内容的提问来源于stack exchange，提问作者Jessica