What is the main approach to solve Maximum Number of Words Found in Sentences?

Iterate through each sentence, count words using spaces, and track the maximum number found in any sentence.

Can I use split method to count words?

Yes, splitting each sentence by spaces gives an array whose length is the word count.

How do I handle single-word sentences?

A single-word sentence has zero spaces, so adding one gives the correct count of one.

What is the time complexity for this problem?

Time complexity is O(n*m) where n is the number of sentences and m is the average sentence length.

Do I need to worry about leading or trailing spaces?

No, constraints guarantee sentences have no leading or trailing spaces and words are separated by exactly one space.

#2114

Easy

auto_awesome数组·string

LeetCode 题解工作台

句子中的最多单词数

一个句子由一些单词以及它们之间的单个空格组成，句子的开头和结尾不会有多余空格。给你一个字符串数组 sentences ，其中 sentences[i] 表示单个句子。请你返回单个句子里单词的最多数目。示例 1：输入： sentences = ["alice and bob l…

数组字符串

题目描述

一个句子由一些单词以及它们之间的单个空格组成，句子的开头和结尾不会有多余空格。

给你一个字符串数组 sentences ，其中 sentences[i] 表示单个句子。

请你返回单个句子里 单词的最多数目 。

示例 1：

输入：sentences = ["alice and bob love leetcode", "i think so too", "this is great thanks very much"]
输出：6
解释：
- 第一个句子 "alice and bob love leetcode" 总共有 5 个单词。
- 第二个句子 "i think so too" 总共有 4 个单词。
- 第三个句子 "this is great thanks very much" 总共有 6 个单词。
所以，单个句子中有最多单词数的是第三个句子，总共有 6 个单词。

示例 2：

输入：sentences = ["please wait", "continue to fight", "continue to win"]
输出：3
解释：可能有多个句子有相同单词数。
这个例子中，第二个句子和第三个句子（加粗斜体）有相同数目的单词数。

提示：

1 <= sentences.length <= 100
1 <= sentences[i].length <= 100
sentences[i] 只包含小写英文字母和 ' ' 。
sentences[i] 的开头和结尾都没有空格。
sentences[i] 中所有单词由单个空格隔开。

lightbulb

解题思路

方法一：空格计数

我们遍历数组 sentences，对于每个句子，我们计算其中的空格数，那么单词数就是空格数加 $1$ 。最后返回最大的单词数即可。

时间复杂度 $O(L)$ ，其中 $L$ 是数组 sentences 中所有字符串的长度之和。空间复杂度 $O(1)$ 。

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class Solution:
    def mostWordsFound(self, sentences: List[str]) -> int:
        return 1 + max(s.count(' ') for s in sentences)

speed

复杂度分析

指标	值
时间	complexity is O(n*m) where n is the number of sentences and m is the average length of a sentence. Space complexity is O(1) if counting manually or O(m) if splitting strings into arrays.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Expecting an array traversal with string manipulation to count words.
question_mark
Looking for edge case handling like single-word sentences and uniform spaces.
question_mark
Evaluates familiarity with split, length, and iterative maximum updates.

warning

常见陷阱

外企场景

error
Counting spaces incorrectly when multiple spaces exist, although constraints prevent this.
error
Forgetting to add one to the space count to get the total words.
error
Assuming empty sentences or trimming is needed, which violates the problem constraints.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return the index of the sentence with the maximum words instead of the count.
arrow_right_alt
Count words ignoring punctuation or non-letter characters in sentences.
arrow_right_alt
Handle sentences with varying spacing and leading/trailing spaces.

help

常见问题

外企场景

继续练习

#2115 从给定原材料中找到所有可以做出的菜 #2109 向字符串添加空格 #2108 找出数组中的第一个回文字符串