查找字符串中最大和最小单词的程序

2025 年 1 月 8 日 | 阅读 5 分钟

说明

在此程序中，我们需要找到字符串中存在的最短和最长的单词。

考虑上面的例子，其中“an”是最短的单词，“extraordinary”是最长的单词。查找最短和最长单词的一种方法是将字符串拆分成单词，然后将每个单词的长度与变量 small 和 large 进行比较。如果一个单词的长度小于 small 的长度，则将该单词存储在 small 中。如果一个单词的长度大于 large 的长度，则将该单词存储在 large 中。

算法

定义一个字符串。
将字符串转换为小写，使其不区分大小写。
在末尾添加一个额外的空格。
现在，遍历字符串直到找到空格，并将这些字符添加到变量 word 中。将每个单词添加到 words 数组中。数组 words 将包含字符串中的所有单词。
使用数组的第一个单词初始化变量 small 和 large。
遍历数组 words，检查单词的长度是否小于 small。如果是，则将该单词存储在 small 中。
如果单词的长度大于 large。如果是，则将该单词存储在 large 中。
最后，显示最短和最长的单词。

解决方案

Python

string = "Hardships often prepare ordinary people for an extraordinary destiny";
word = "";
words = [];
 
#Add extra space after string to get the last word in the given string
string = string + " ";
 
for i in range(0, len(string)):
    #Split the string into words
    if(string[i] != ' '):
        word = word + string[i];
    else:
        #Add word to array words
        words.append(word);
        #Make word an empty string
        word = "";
        
#Initialize small and large with first word in the string
small = large = words[0];
 
#Determine smallest and largest word in the string
for k in range(0, len(words)):
    
    #If length of small is greater than any word present in the string
    #Store value of word into small
    if(len(small) > len(words[k])):
        small = words[k];
        
    #If length of large is less than any word present in the string
    #Store value of word into large
    if(len(large) < len(words[k])):
        large = words[k];
 
print("Smallest word: " + small);
print("Largest word: " + large);

输出

Smallest word: an
Largest word: extraordinary

C

#include <stdio.h>
#include <string.h>
 
int main()
{   
    char string[] = "Hardships often prepare ordinary people for an extraordinary destiny";
    char words[100][100], small[100], large[100];
    int i = 0, j = 0, k, length;
    
    //Split the string into words such that each row of array words represents a word
    for(k=0; string[k]!='\0'; k++){
        
        //Here, i represents row and j represents column of two-dimensional array words 
        if(string[k] != ' ' && string[k] != '\0'){
            words[i][j++] = string[k];
        }
        else{
            words[i][j] = '\0';
            //Increment row count to store new word
            i++;
            //Set column count to 0
            j = 0;
        }
    }
    
    //Store row count in variable length
    length = i + 1;
    
    //Initialize small and large with first word in the string
    strcpy(small, words[0]);
    strcpy(large, words[0]);
    
    //Determine smallest and largest word in the string
    for(k = 0; k < length; k++){
        
        //If length of small is greater than any word present in the string
        //Store value of word into small
        if(strlen(small) > strlen(words[k])){
            strcpy(small, words[k]);
        }
        //If length of large is less than any word present in the string
        //Store value of word into large
        if(strlen(large) < strlen(words[k]))
            strcpy(large, words[k]);
    }
    
    printf("Smallest word: %s\n", small);
    printf("Largest word: %s", large);
    
    return 0;
}

输出

Smallest word: an
Largest word: extraordinary

JAVA

public class SmallestLargestWord
{
    public static void main(String[] args){ 
        String string = "Hardships often prepare ordinary people for an extraordinary destiny";
        String word = "", small = "", large="";
        String[] words = new String[100];
        int length = 0;
        
        //Add extra space after string to get the last word in the given string
        string = string + " ";
        
        for(int i = 0; i < string.length(); i++){
            //Split the string into words
            if(string.charAt(i) != ' '){
                word = word + string.charAt(i);
            }
            else{
                //Add word to array words
                words[length] = word;
                //Increment length
                length++;
                //Make word an empty string
                word = "";
            }
        }
        
        //Initialize small and large with first word in the string
        small = large = words[0];
        
        //Determine smallest and largest word in the string
        for(int k = 0; k < length; k++){
            
            //If length of small is greater than any word present in the string
            //Store value of word into small
            if(small.length() > words[k].length())
                small = words[k];
 
            //If length of large is less than any word present in the string
            //Store value of word into large
            if(large.length() < words[k].length())
                large = words[k];
        }
        
        System.out.println("Smallest word: " + small);
        System.out.println("Largest word: " + large);
    }
}

输出

Smallest word: an
Largest word: extraordinary

C#

using System;
                    
public class SmallestLargestWord
{
    public static void Main()
    {
        String string1 = "Hardships often prepare ordinary people for an extraordinary destiny";
        String word = "", small = "", large="";
        String[] words = new String[100];
        int length = 0;
        
        //Add extra space after string1 to get the last word in the given string
        string1 = string1 + " ";
        
        for(int i = 0; i < string1.Length; i++){
            //Split the string into words
            if(string1[i] != ' '){
                word = word + string1[i];
            }
            else{
                //Add word to array words
                words[length] = word;
                //Increment length
                length++;
                //Make word an empty string
                word = "";
            }
        }
        
        //Initialize small and large with first word in the string
        small = large = words[0];
        
        //Determine smallest and largest word in the string
        for(int k = 0; k < length; k++){
            
            //If length of small is greater than any word present in the string
            //Store value of word into small
            if(small.Length > words[k].Length)
                small = words[k];
 
            //If length of large is less than any word present in the string
            //Store value of word into large
            if(large.Length < words[k].Length)
                large = words[k];
        }
        
        Console.WriteLine("Smallest word: " + small);
        Console.WriteLine("Largest word: " + large);
    }
}

输出

Smallest word: an
Largest word: extraordinary

PHP

<!DOCTYPE html>
<html>
<body>
<?php
$string = "Hardships often prepare ordinary people for an extraordinary destiny";
$word = "";
$words = array();
 
//Add extra space after string to get the last word in the given string
$string = $string . " ";
 
for($i = 0; $i < strlen($string); $i++){
    //Split the string into words
    if($string[$i] != ' '){
        $word = $word . $string[$i];
    }
    else{
        //Add word to string array words
        array_push($words, $word);
        //Make word an empty string
        $word = "";
    }
}
 
//Initialize small and large with first word in the string
$small = $large = $words[0];
 
//Determine smallest and largest word in the string
for($k = 0; $k < count($words); $k++){
    
    //If length of small is greater than any word present in the string
    //Store value of word into small
    if(strlen($small) > strlen($words[$k]))
        $small = $words[$k];
 
    //If length of large is less than any word present in the string
    //Store value of word into large
    if(strlen($large) < strlen($words[$k]))
        $large = $words[$k];
}
 
print("Smallest word: " . $small);
print("<br>Largest word: " . $large);
?>
</body>
</html>

输出

 Smallest word: an
Largest word: extraordinary

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查找字符串中最大和最小单词的程序

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查找字符串中最大和最小单词的程序

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