What is the main approach for solving the Length of the Longest Alphabetical Continuous Substring problem?

The key approach is to iterate through the string, comparing adjacent characters, and keep track of the longest continuous substring that appears in alphabetical order.

How do I handle cases where the string is too short for a meaningful alphabetical substring?

In such cases, a single-character string would have the longest alphabetical substring of length 1. Ensure that this edge case is handled.

What happens when no characters form an alphabetical sequence?

If no characters form a continuous alphabetical sequence, each individual character is considered its own substring, and the length of the longest substring will be 1.

How does the problem relate to the string-driven solution strategy?

This problem is a prime example of using a string-driven approach where you manipulate and process a string character by character to identify patterns or substrings.

What is the time complexity of the solution for the Length of the Longest Alphabetical Continuous Substring problem?

The time complexity of the optimal solution is O(n), where n is the length of the string, as we process each character once.

#2414

Medium

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

最长的字母序连续子字符串的长度

字母序连续字符串是由字母表中连续字母组成的字符串。换句话说，字符串 "abcdefghijklmnopqrstuvwxyz" 的任意子字符串都是字母序连续字符串。例如， "abc" 是一个字母序连续字符串，而 "acb" 和 "za" 不是。给你一个仅由小写英文字母组成的字符串 s ，返回…

字符串

题目描述

字母序连续字符串 是由字母表中连续字母组成的字符串。换句话说，字符串 "abcdefghijklmnopqrstuvwxyz" 的任意子字符串都是 字母序连续字符串 。

例如，"abc" 是一个字母序连续字符串，而 "acb" 和 "za" 不是。

给你一个仅由小写英文字母组成的字符串 s ，返回其最长的字母序连续子字符串的长度。

示例 1：

输入：s = "abacaba"
输出：2
解释：共有 4 个不同的字母序连续子字符串 "a"、"b"、"c" 和 "ab" 。
"ab" 是最长的字母序连续子字符串。

示例 2：

输入：s = "abcde"
输出：5
解释："abcde" 是最长的字母序连续子字符串。

提示：

1 <= s.length <= 10⁵
s 由小写英文字母组成

lightbulb

解题思路

方法一：一次遍历

我们可以遍历字符串 $s$ ，用一个变量 $\textit{ans}$ 记录最长的字母序连续子字符串的长度，用另一个变量 $\textit{cnt}$ 记录当前连续子字符串的长度。初始时 $\textit{ans} = \textit{cnt} = 1$ 。

接下来，我们从下标为 $1$ 的字符开始遍历字符串 $s$ ，对于每个字符 $s[i]$ ，如果 $s[i] - s[i - 1] = 1$ ，则说明当前字符和前一个字符是连续的，此时 $\textit{cnt} = \textit{cnt} + 1$ ，并更新 $\textit{ans} = \max(\textit{ans}, \textit{cnt})$ ；否则，说明当前字符和前一个字符不连续，此时 $\textit{cnt} = 1$ 。

最终返回 $\textit{ans}$ 即可。

时间复杂度 $O(n)$ ，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def longestContinuousSubstring(self, s: str) -> int:
        ans = cnt = 1
        for x, y in pairwise(map(ord, s)):
            if y - x == 1:
                cnt += 1
                ans = max(ans, cnt)
            else:
                cnt = 1
        return ans

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Ability to track the longest substring efficiently
question_mark
Effective use of variables to maintain substring length
question_mark
Clear understanding of edge cases and constraints

warning

常见陷阱

外企场景

error
Overcomplicating the problem with unnecessary nested loops
error
Not properly handling edge cases such as strings of length one or no alphabetical sequences
error
Failure to update the longest substring correctly during iteration

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider different ways to scan through the string, such as using sliding window techniques
arrow_right_alt
Extend the problem to handle uppercase letters as well
arrow_right_alt
Modify the problem to return the actual substring instead of just its length

help

常见问题

外企场景

继续练习

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