0

此代码从用户那里获取一个输入字符串，其中可能包含特殊字符。然后我们输出字母表中缺少的字母。例如，“the quick brown fox jumped over the lazy dog”将返回一个空字符串，但“ZYXW, vu TSR Ponm lkj ihgfd CBA”。会缺少字母“eq”。

目前，我的程序正在返回整个字母表，而不仅仅是丢失的字符。


import java.io.*;

import org.apache.commons.lang3.*;

public class QuickBrownFox {

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String s = reader.readLine();
        s = s.toUpperCase();
        String[] arr = { "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S",
                "T", "U", "V", "W", "X", "Y", "Z" };
        String chars_only = "";
        for (int i = 0; i < s.length(); i++) {
            for (int j = 0; j < arr.length; j++) {
                if (s.substring(i, i + 1).equals(arr[j])) {
                    chars_only += s.substring(i, i + 1);
                }
            }
        }
        System.out.println(chars_only); // now we have a string of only alphabet letters
        String missing = "";
        for (int j = 0; j < arr.length; j++) {
            if (StringUtils.contains(arr[j], chars_only) == false) { // alphabet letter not found
                missing += arr[j];
            }
        }
        missing = missing.toLowerCase();
        System.out.println("missing letters: " + missing);
    }
}
4

2 回答 2

0

一种选择是使用Set用给定的初始化alphabet-chars。然后从set任何chars包含在input-string. 余数将反映缺失chars。

import java.util.Collections;
import java.util.HashSet;
import java.util.Set;

public class TestAlphabet {
    public static void main(String[] args) {
        
        Set<Character> set = new HashSet<Character>();
        Character[] arr = { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S',
                'T', 'U', 'V', 'W', 'X', 'Y', 'Z' };
        Collections.addAll(set, arr);
        String s1="abc";
        System.out.println("input:"+s1);
        new TestAlphabet().new Missing().getMissing(set, s1).forEach(System.out::print);
        System.out.println("\n");
        String s2="the quick brown fox jumped over the lazy dog";
        System.out.println("input:"+s2);
        new TestAlphabet().new Missing().getMissing(set, s2).forEach(System.out::print);
        System.out.println("\n");
        String s3="ZYXW, vu TSR Ponm lkj ihgfd CBA.";
        System.out.println("input:"+s3);
        new TestAlphabet().new Missing().getMissing(set, s3).forEach(System.out::print);
    }
    class Missing
    {
        public Set<Character> getMissing(Set<Character> orig, String s)
        {
            //keep origin
            Set<Character> set = new HashSet<Character>();
            set.addAll(orig);
            s=s.toUpperCase();
            for(int i=0;i<s.length();i++)
            {
                Character c=s.charAt(i);
                set.remove(c);
            }
            return set;
        }
    }
}

输出

input:abc
DEFGHIJKLMNOPQRSTUVWXYZ

input:the quick brown fox jumped over the lazy dog
S

input:ZYXW, vu TSR Ponm lkj ihgfd CBA.
EQ
于 2021-02-06T00:18:06.213 回答
0

StringUtils.contains(arr[j], chars_only)检查字母 ( arr[j]) 是否包含整个字符串chars_only，这种情况永远不会出现（除非您只输入一个字符）。你想要的是!chars_only.contains(arr[j])（不需要使用StringUtils.contains）。这将检查输入的字母是否不包含第jth 个字母。或者，您可以保留StringUtils.contains，只需交换参数：!StringUtils.contains(chars_only, arr[j]).

顺便说一句，您可以使用流来减少代码量：

import java.util.Arrays;
import java.util.List;

import java.io.*;
import java.util.stream.Collectors;

public class QuickBrownFox {
  public static void main(String[] args) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    String s = reader.readLine().toUpperCase();

    List<String> alphabet = Arrays.asList("A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S",
      "T", "U", "V", "W", "X", "Y", "Z");
    String missing = alphabet.stream()
      .filter(letter -> !s.contains(letter))
      .collect(Collectors.joining())
      .toLowerCase();

    System.out.println("missing letters: " + missing);
  }
}
于 2021-02-05T23:38:49.593 回答