[leetcode] 3. Longest Substring Without Repeating Characters @ python

本文介绍了解决LeetCode上“最长无重复字符子串”问题的一种高效算法。通过遍历字符串并使用字典记录字符及其索引,实现O(n)时间复杂度下寻找最长无重复子串长度。示例包括'abcabcbb'、'bbbbb'和'pwwkew'等。

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

原题

https://leetcode.com/problems/longest-substring-without-repeating-characters/
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.

解题思路

遍历字符串, 用字典来储存字符和索引, 如果字符重复了, 更新subtring的起点start, 每次遍历后, 更新result的值. 这里start应该只能增加不能减少, 否则特殊情况时, 例如’abba’ 会出错.

时间复杂度: O(n)
空间复杂度: O(1)

代码

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        # use start to record the start index of substring
        start, res = 0, 0
        seen = {}
        for i, ch in enumerate(s):
            if ch in seen:
                # start cannot decrease
                start = max(start, seen[ch] + 1)
            seen[ch] = i
            res = max(res, i-start+1)
        return res
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