[python] Leetcode 3. Longest Substring Without Repeating Characters

最新推荐文章于 2025-08-13 21:50:19 发布

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最新推荐文章于 2025-08-13 21:50:19 发布

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分类专栏： Leetcode 文章标签： leetcode python 面试刷题

本文链接：https://blog.youkuaiyun.com/qq_31248309/article/details/86665195

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本文探讨了在给定字符串中寻找最长无重复字符子串的两种方法。通过使用字典跟踪字符最后出现的位置和利用集合来判断字符是否已存在，实现了高效的解决方案。示例包括“abcabcbb”、“bbbbb”和“pwwkew”，详细解释了如何找到正确答案。

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题目描述

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.

代码实现

# 两种方法的left和right存的都是不重复最长字串
class Solution(object):
    def lengthOfLongestSubstring(self, s):  
     # 方法一，wordidx存的是当前时刻出现的所有词的最新index，这个index可能是在left前（例：pwwped的第2个p）
        """
        :type s: str
        :rtype: int
        """
        res=0
        left,right = 0,0
        wordidx = dict()
        for right in range(len(s)):
            if s[right] in wordidx:
                left = max(left, wordidx[s[right]]+1)
            wordidx[s[right]]=right
            res = max(res, right-left+1)
        return res
    
    def lengthOfLongestSubstring(self, s):   
    # 方法二，set与left和right间字串对应，存的是当前位置的最长字串的字符，与方法一在确定left边界时有区别
        """
        :type s: str
        :rtype: int
        """
        res=0
        left,right = 0,0
        words = set()
        for right in range(len(s)):
            while s[right] in words:
                words.remove(s[left])
                left+=1
            words.add(s[right])
            res = max(res,len(words))        
        return res