What happens if a string has multiple consecutive separators?

Consecutive separators produce empty strings when split, which should be filtered out so they are not included in the final array.

Can the separator appear at the start or end of a string?

Yes, but any resulting empty substrings from leading or trailing separators must be ignored in the output.

Is the order of strings important in the output?

Yes, the resulting array should maintain the original word order and the sequence of splits within each word.

How does this relate to the Array plus String pattern?

This problem combines array iteration with string manipulation, a hallmark of the Array plus String pattern, focusing on split operations and result aggregation.

What if the array contains only separators?

If all strings only contain separators, splitting them results in empty strings, so the final returned array will be empty.

#2788

Easy

auto_awesome数组·string

LeetCode 题解工作台

按分隔符拆分字符串

给你一个字符串数组 words 和一个字符 separator ，请你按 separator 拆分 words 中的每个字符串。返回一个由拆分后的新字符串组成的字符串数组，不包括空字符串。注意 separator 用于决定拆分发生的位置，但它不包含在结果字符串中。拆分可能形成两个以上的字符…

数组字符串

题目描述

给你一个字符串数组 words 和一个字符 separator ，请你按 separator 拆分 words 中的每个字符串。

返回一个由拆分后的新字符串组成的字符串数组，不包括空字符串 。

注意

separator 用于决定拆分发生的位置，但它不包含在结果字符串中。
拆分可能形成两个以上的字符串。
结果字符串必须保持初始相同的先后顺序。

示例 1：

输入：words = ["one.two.three","four.five","six"], separator = "."
输出：["one","two","three","four","five","six"]
解释：在本示例中，我们进行下述拆分：

"one.two.three" 拆分为 "one", "two", "three"
"four.five" 拆分为 "four", "five"
"six" 拆分为 "six" 

因此，结果数组为 ["one","two","three","four","five","six"] 。

示例 2：

输入：words = ["$easy$","$problem$"], separator = "$"
输出：["easy","problem"]
解释：在本示例中，我们进行下述拆分：

"$easy$" 拆分为 "easy"（不包括空字符串）
"$problem$" 拆分为 "problem"（不包括空字符串）

因此，结果数组为 ["easy","problem"] 。

示例 3：

输入：words = ["|||"], separator = "|"
输出：[]
解释：在本示例中，"|||" 的拆分结果将只包含一些空字符串，所以我们返回一个空数组 [] 。

提示：

1 <= words.length <= 100
1 <= words[i].length <= 20
words[i] 中的字符要么是小写英文字母，要么就是字符串 ".,|$#@" 中的字符（不包括引号）
separator 是字符串 ".,|$#@" 中的某个字符（不包括引号）

lightbulb

解题思路

方法一：模拟

我们遍历字符串数组 $words$ ，对于每个字符串 $w$ ，我们使用 separator 作为分隔符进行拆分，如果拆分后的字符串不为空，则将其加入答案数组。

时间复杂度 $O(n \times m)$ ，空间复杂度 $O(m)$ ，其中 $n$ 是字符串数组 $words$ 的长度，而 $m$ 是字符串数组 $words$ 中字符串的最大长度。

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class Solution:
    def splitWordsBySeparator(self, words: List[str], separator: str) -> List[str]:
        return [s for w in words for s in w.split(separator) if s]

speed

复杂度分析

指标	值
时间	complexity is O(n * m) where n is the number of strings and m is the average string length, due to iterating and splitting each string. Space complexity is O(total number of resulting substrings) as we store all non-empty parts.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for handling consecutive separators correctly.
question_mark
Ensure empty strings are not included in the result.
question_mark
Check if the solution maintains the original order of words and splits.

warning

常见陷阱

外企场景

error
Including empty strings after splitting at consecutive separators.
error
Not preserving the order of strings after flattening all splits.
error
Using complex nested loops instead of simple iteration and split.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Split strings by multiple different separator characters instead of one.
arrow_right_alt
Return only substrings of a certain minimum length after splitting.
arrow_right_alt
Split a single long string multiple times using an array of separators.

help

常见问题

外企场景

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