从一个给定的句子中找出一个单词作为前缀
给定一个代表一个句子的字符串 S 和另一个字符串单词，任务是从 S 中找到这个单词，这个单词有字符串单词作为前缀。如果字符串中没有这样的单词，打印 -1 。
示例:
输入: S =“欢迎来到 Geeksforgeeks”，word =“Gee” 输出: Geeksforgeeks 解释: 句子中的“Geeksforgeeks”一词有前缀“Gee”。
输入: s=“竞技编程”，word =“kdflk” 输出: -1 解释: 字符串中没有单词以“kdflk”为前缀。
方法:按照以下步骤查找哪个单词有给定的前缀:
- 使用字符串流从句子中提取单词，并将它们存储在字符串向量中。
- 现在，遍历数组并检查哪个单词包含给定的单词作为它自己的前缀。
- 如果发现任何一个单词都是正确的，那么打印这个单词。否则，如果没有找到这样的字，打印 -1 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the position
// of the string having word as prefix
string isPrefixOfWord(string sentence,
string Word)
{
stringstream ss(sentence);
// Initialize a vector
vector<string> v;
string temp;
// Extract words from the sentence
while (ss >> temp) {
v.push_back(temp);
}
// Traverse each word
for (int i = 0; i < v.size(); i++) {
// Traverse the characters of word
for (int j = 0; j < v[i].size(); j++) {
// If prefix does not match
if (v[i][j] != Word[j])
break;
// Otherwise
else if (j == Word.size() - 1)
// Return the word
return v[i];
}
}
// Return -1 if not present
return "-1";
}
// Driver code
int main()
{
string s = "Welcome to Geeksforgeeks";
string word = "Gee";
cout << isPrefixOfWord(s, word) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
import java.io.*;
import java.lang.Math;
class GFG{
// Function to find the position
// of the string having word as prefix
static String isPrefixOfWord(String sentence,
String Word)
{
String a[] = sentence.split(" ");
// Initialize an ArrayList
ArrayList<String> v = new ArrayList<>();
// Extract words from the sentence
for(String e : a)
v.add(e);
// Traverse each word
for(int i = 0; i < v.size(); i++)
{
// Traverse the characters of word
for(int j = 0; j < v.get(i).length(); j++)
{
// If prefix does not match
if (v.get(i).charAt(j) != Word.charAt(j))
break;
// Otherwise
else if (j == Word.length() - 1)
// Return the word
return v.get(i);
}
}
// Return -1 if not present
return "-1";
}
// Driver code
public static void main(final String[] args)
{
String s = "Welcome to Geeksforgeeks";
String word = "Gee";
System.out.println(isPrefixOfWord(s, word));
}
}
// This code is contributed by bikram2001jha
Python 3
# Python3 program for the
# above approach
# Function to find the
# position of the string
# having word as prefix
def isPrefixOfWord(sentence, word):
a=sentence.split(" ")
# Initialize an List
v = []
# Extract words from
# the sentence
for i in a:
v.append(i)
# Traverse each word
for i in range(len(v)):
# Traverse the characters of word
for j in range(len(v[i])):
# If prefix does not match
if(v[i][j] != word[j]):
break
# Otherwise
elif(j == len(word) - 1):
# Return the word
return v[i]
# Return -1 if not present
return -1
# Driver code
s = "Welcome to Geeksforgeeks"
word = "Gee"
print(isPrefixOfWord(s, word))
# This code is contributed by avanitrachhadiya2155
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the position
// of the string having word as prefix
static String isPrefixOfWord(String sentence,
String Word)
{
String []a = sentence.Split(' ');
// Initialize an List
List<String> v = new List<String>();
// Extract words from the sentence
foreach(String e in a)
v.Add(e);
// Traverse each word
for(int i = 0; i < v.Count; i++)
{
// Traverse the characters of word
for(int j = 0; j < v[i].Length; j++)
{
// If prefix does not match
if (v[i][j] != Word[j])
break;
// Otherwise
else if (j == Word.Length - 1)
// Return the word
return v[i];
}
}
// Return -1 if not present
return "-1";
}
// Driver code
public static void Main(String[] args)
{
String s = "Welcome to Geeksforgeeks";
String word = "Gee";
Console.WriteLine(isPrefixOfWord(s, word));
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
//Javascript program for the above approach
// Function to find the position
// of the string having word as prefix
function isPrefixOfWord(sentence, Word)
{
var a = sentence.split(" ");
// Initialize an ArrayList
var v = [];
// Extract words from the sentence
//for(String e : a)
for(var i=0;i<a.length;i++)
{
v.push(a[i]);
}
// Traverse each word
for(var i = 0; i < v.length; i++)
{
// Traverse the characters of word
for(var j = 0; j < v[i].length; j++)
{
// If prefix does not match
if (v[i].charAt(j) != Word[j])
break;
// Otherwise
else if (j == Word.length- 1)
// Return the word
return v[i];
}
}
// Return -1 if not present
return "-1";
}
var s = "Welcome to Geeksforgeeks";
var word = "Gee";
document.write(isPrefixOfWord(s, word));
//This code in contributed by SoumikMondal
</script>
Output:
Geeksforgeeks
时间复杂度: O(L)，其中 L 表示字符串 S 的长度 辅助空间: O(L)
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