What is the main approach for Sorting the Sentence on LeetCode?

Split words, extract numeric positions, sort or place them according to indices, and remove numbers to rebuild the sentence.

Can this problem be solved without explicit sorting?

Yes, since the number of words is limited, you can directly place each word in a pre-sized array at its index.

How do I handle single-word or two-word sentences?

Even with minimal words, apply the same extraction and placement logic; the solution naturally handles small cases.

Are there any edge cases for index numbers?

Indices are 1-based and at most 9, so verify parsing extracts digits correctly without assuming multi-digit numbers.

Does the Sorting plus String pattern generalize to other problems?

Yes, this pattern of extracting keys from strings and ordering elements is common for similar string sorting tasks.

#1859

Easy

auto_awesomestring·结合·排序

LeetCode 题解工作台

将句子排序

一个句子指的是一个序列的单词用单个空格连接起来，且开头和结尾没有任何空格。每个单词都只包含小写或大写英文字母。我们可以给一个句子添加从 1 开始的单词位置索引，并且将句子中所有单词打乱顺序。比方说，句子 "This is a sentence" 可以被打乱顺序得到 "sentence…

字符串排序

题目描述

一个句子指的是一个序列的单词用单个空格连接起来，且开头和结尾没有任何空格。每个单词都只包含小写或大写英文字母。

我们可以给一个句子添加 从 1 开始的单词位置索引 ，并且将句子中所有单词 打乱顺序 。

比方说，句子 "This is a sentence" 可以被打乱顺序得到 "sentence4 a3 is2 This1" 或者 "is2 sentence4 This1 a3" 。

给你一个 打乱顺序 的句子 s ，它包含的单词不超过 9 个，请你重新构造并得到原本顺序的句子。

示例 1：

输入：s = "is2 sentence4 This1 a3"
输出："This is a sentence"
解释：将 s 中的单词按照初始位置排序，得到 "This1 is2 a3 sentence4" ，然后删除数字。

示例 2：

输入：s = "Myself2 Me1 I4 and3"
输出："Me Myself and I"
解释：将 s 中的单词按照初始位置排序，得到 "Me1 Myself2 and3 I4" ，然后删除数字。

提示：

2 <= s.length <= 200
s 只包含小写和大写英文字母、空格以及从 1 到 9 的数字。
s 中单词数目为 1 到 9 个。
s 中的单词由单个空格分隔。
s 不包含任何前导或者后缀空格。

lightbulb

解题思路

方法一：字符串分割

我们先将字符串 $s$ 按照空格分割，得到字符串数组 $\textit{ws}$ ，然后遍历数组 $\textit{ws}$ ，将每个单词的最后一个字符减去字符 '1'，得到的结果作为单词的索引，将单词的前缀作为单词的内容，最后将单词按照索引顺序拼接起来即可。

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

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class Solution:
    def sortSentence(self, s: str) -> str:
        ws = [(w[:-1], int(w[-1])) for w in s.split()]
        ws.sort(key=lambda x: x[1])
        return ' '.join(w for w, _ in ws)

speed

复杂度分析

指标	值
时间	complexity is O(n) for parsing and O(n log n) if sorting by indices, but since n <= 9, the practical runtime is constant. Space complexity is O(n) for storing words and their positions.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Check if the candidate efficiently extracts numeric indices from words without extra string manipulations.
question_mark
Watch for correct placement of words in their original positions, especially when input is already partially ordered.
question_mark
Confirm candidate handles edge cases with minimum or maximum number of words correctly.

warning

常见陷阱

外企场景

error
Forgetting to remove the numeric suffix after sorting, resulting in incorrect output.
error
Using unstable sorting which can mix up words with similar numeric positions.
error
Incorrectly parsing the string, especially splitting or indexing errors on words.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Reconstruct sentences with multi-digit indices requiring robust parsing logic.
arrow_right_alt
Handle shuffled sentences with punctuation attached to words, needing clean separation before sorting.
arrow_right_alt
Optimize in-place sorting using index mapping instead of extra arrays for space efficiency.

help

常见问题

外企场景

继续练习

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