How do I handle a string with only one word in "Rearrange Spaces Between Words"?

All spaces should be appended at the end since no spaces can exist between words, ensuring the string length remains the same.

What is the main pattern for solving this problem efficiently?

Use a string-driven solution strategy: count words and spaces, then divide spaces evenly between words.

How do I deal with leftover spaces after equal distribution?

Append any leftover spaces to the end of the string so that the output matches the original length.

Can this approach handle leading and trailing spaces?

Yes, the algorithm counts all spaces regardless of position and redistributes them according to the problem rules.

Why is integer division used in this problem?

Integer division determines the maximum number of spaces that can be evenly placed between words while separating leftover spaces for the end.

#1592

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

重新排列单词间的空格

给你一个字符串 text ，该字符串由若干被空格包围的单词组成。每个单词由一个或者多个小写英文字母组成，并且两个单词之间至少存在一个空格。题目测试用例保证 text 至少包含一个单词。请你重新排列空格，使每对相邻单词之间的空格数目都相等，并尽可能最大化该数目。如果不能重新平均分配所有空格…

字符串

题目描述

给你一个字符串 text ，该字符串由若干被空格包围的单词组成。每个单词由一个或者多个小写英文字母组成，并且两个单词之间至少存在一个空格。题目测试用例保证 text 至少包含一个单词 。

请你重新排列空格，使每对相邻单词之间的空格数目都相等，并尽可能 最大化 该数目。如果不能重新平均分配所有空格，请 将多余的空格放置在字符串末尾 ，这也意味着返回的字符串应当与原 text 字符串的长度相等。

返回 重新排列空格后的字符串 。

示例 1：

输入：text = "  this   is  a sentence "
输出："this   is   a   sentence"
解释：总共有 9 个空格和 4 个单词。可以将 9 个空格平均分配到相邻单词之间，相邻单词间空格数为：9 / (4-1) = 3 个。

示例 2：

输入：text = " practice   makes   perfect"
输出："practice   makes   perfect "
解释：总共有 7 个空格和 3 个单词。7 / (3-1) = 3 个空格加上 1 个多余的空格。多余的空格需要放在字符串的末尾。

示例 3：

输入：text = "hello   world"
输出："hello   world"

示例 4：

输入：text = "  walks  udp package   into  bar a"
输出："walks  udp  package  into  bar  a "

示例 5：

输入：text = "a"
输出："a"

提示：

1 <= text.length <= 100
text 由小写英文字母和 ' ' 组成
text 中至少包含一个单词

lightbulb

解题思路

方法一：字符串模拟

我们先统计字符串 $\textit{text}$ 中的空格数，记为 $\textit{spaces}$ 。将 $\textit{text}$ 按空格分割成字符串数组 $\textit{words}$ 。然后计算相邻字符串之间需要拼接的空格数，进行拼接。最后将剩余的空格拼接在末尾。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ ，其中 $n$ 表示字符串 $\textit{text}$ 的长度。

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class Solution:
    def reorderSpaces(self, text: str) -> str:
        spaces = text.count(" ")
        words = text.split()
        if len(words) == 1:
            return words[0] + " " * spaces
        cnt, mod = divmod(spaces, len(words) - 1)
        return (" " * cnt).join(words) + " " * mod

speed

复杂度分析

指标	值
时间	complexity is O(n) to scan the string and count words and spaces. Space complexity is O(n) for storing words and building the output string.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Looking for efficient string parsing without extra passes over the text.
question_mark
Expect candidate to handle cases with a single word where no spaces can be distributed between words.
question_mark
Watch for correct placement of leftover spaces at the end to preserve original string length.

warning

常见陷阱

外企场景

error
Forgetting to handle the single-word edge case, which requires all spaces at the end.
error
Incorrectly dividing spaces when words are fewer than two, leading to division by zero.
error
Appending leftover spaces incorrectly, breaking the original string length constraint.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Redistribute spaces but preserve punctuation attached to words.
arrow_right_alt
Maximize spacing while also aligning words to the right side of the string.
arrow_right_alt
Redistribute spaces evenly for multi-line strings where each line must maintain its original length.

help

常见问题

外企场景

继续练习

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