【LeetCode】3. Longest Substring Without Repeating Characters 解题报告(Python)

博客给出题目,要求找出给定字符串中无重复字符子串的长度。还展示了一段Python代码,该代码根据right指向的字符是否出现在集合中进行循环来解决此问题,并提供了参考资料链接。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

 

题目描述:

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

 

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", which the length is 3.


Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.


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.

下面的python代码是根据right指向的字符是否出现在set中而反复的进行循环,代码如下:

class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        left, right = 0, 0
        chars = set()
        res = 0
        while left < len(s) and right < len(s):
            if s[right] in chars:
                if s[left] in chars:
                    chars.remove(s[left])
                left += 1
            else:
                chars.add(s[right])
                right += 1
                res = max(res, len(chars))
        return res

参考资料:https://blog.youkuaiyun.com/fuxuemingzhu/article/details/82022530

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值