What pattern does this problem follow?

This problem uses the two-pointer scanning with invariant tracking pattern, focusing on efficient prefix detection within a sentence.

How do I handle multiple words matching the prefix?

Return only the 1-based index of the first word that matches the searchWord prefix to satisfy minimal index requirement.

Can I use built-in split functions?

Yes, but using two-pointer scanning reduces extra memory allocation and aligns with the two-pointer invariant approach.

What is the expected time complexity?

The solution should run in O(n + m) time where n is sentence length and m is searchWord length, scanning each character once.

How to implement prefix comparison correctly?

Compare only the first k characters of each word with the searchWord of length k to efficiently detect prefix matches without extra checks.

#1455

Easy

auto_awesome双·指针·invariant

LeetCode 题解工作台

检查单词是否为句中其他单词的前缀

给你一个字符串 sentence 作为句子并指定检索词为 searchWord ，其中句子由若干用单个空格分隔的单词组成。请你检查检索词 searchWord 是否为句子 sentence 中任意单词的前缀。如果 searchWord 是某一个单词的前缀，则返回句子 sentence 中该单词…

双指针字符串字符串匹配

题目描述

给你一个字符串 sentence 作为句子并指定检索词为 searchWord ，其中句子由若干用 单个空格 分隔的单词组成。请你检查检索词 searchWord 是否为句子 sentence 中任意单词的前缀。

如果 searchWord 是某一个单词的前缀，则返回句子 sentence 中该单词所对应的下标（下标从 1 开始）。如果 searchWord 是多个单词的前缀，则返回匹配的第一个单词的下标（最小下标）。如果 searchWord 不是任何单词的前缀，则返回 -1 。

字符串 s 的前缀是 s 的任何前导连续子字符串。

示例 1：

输入：sentence = "i love eating burger", searchWord = "burg"
输出：4
解释："burg" 是 "burger" 的前缀，而 "burger" 是句子中第 4 个单词。

示例 2：

输入：sentence = "this problem is an easy problem", searchWord = "pro"
输出：2
解释："pro" 是 "problem" 的前缀，而 "problem" 是句子中第 2 个也是第 6 个单词，但是应该返回最小下标 2 。

示例 3：

输入：sentence = "i am tired", searchWord = "you"
输出：-1
解释："you" 不是句子中任何单词的前缀。

提示：

1 <= sentence.length <= 100
1 <= searchWord.length <= 10
sentence 由小写英文字母和空格组成。
searchWord 由小写英文字母组成。

lightbulb

解题思路

方法一：字符串分割

我们将 $\textit{sentence}$ 按空格分割为 $\textit{words}$ ，然后遍历 $\textit{words}$ ，检查 $\textit{words}[i]$ 是否是 $\textit{searchWord}$ 的前缀，是则返回 $i+1$ 。若遍历结束，所有单词都不满足，返回 $-1$ 。

时间复杂度 $O(m \times n)$ ，空间复杂度 $O(m)$ 。其中 $m$ 和 $n$ 分别是 $\textit{sentence}$ 和 $\textit{searchWord}$ 的长度。

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class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        for i, s in enumerate(sentence.split(), 1):
            if s.startswith(searchWord):
                return i
        return -1

speed

复杂度分析

指标	值
时间	complexity is O(n + m) where n is the sentence length and m is the searchWord length, since each character is visited once in two-pointer scanning. Space complexity is O(n) for storing words or temporary slices during scanning, ensuring linear memory usage.
空间	O(n)

psychology

面试官常问的追问

外企场景

question_mark
Candidate attempts direct string splitting instead of incremental scanning.
question_mark
Candidate compares full words instead of only prefix length.
question_mark
Candidate misses 1-based indexing requirement for the returned word.

warning

常见陷阱

外企场景

error
Checking all characters of each word instead of only the prefix length.
error
Returning index of last matching word instead of first minimal index.
error
Ignoring edge cases where searchWord does not match any word, causing incorrect defaults.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return all indices of words where searchWord is a prefix, instead of first match.
arrow_right_alt
Allow uppercase letters and ignore case during prefix comparison.
arrow_right_alt
Find the longest matching prefix instead of the first word match.

help

常见问题

外企场景

继续练习

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