Question

我试图制造一个阵列,除空间外,删除重复之处(“”)。我看到了一些在线方法,但我不想复制。我尝试了许多方法,但我不理解我的方法为什么不奏效。

这是我所尝试的:

在第一个方法中,该方法是接收投入、打电话、制作一个阵列清单、通过插图打字,同时在阵列清单中增加每个特性,第二次通过清单将每个特性与第一个指数的特性进行比较,同时检查这些特性是否平等,然后从阵列清单中删除。

问题似乎是“在平等后将其从阵列清单中删除”:

if(character.get(i).equals(character.get(k))) {
    character.remove(arr.charAt(i));                
}

第二次

for(int k = 0;k < characters.length;k++) {
    if(characters[i]==characters[k]) {
        System.out.print(character);                        
        character.remove(characters[i]);                    
    }

import java.util.*;

public class NoRepeats {
    
    public static void main(String args[]) {
        System.out.print("Please enter a phrase: ");
        Scanner scanner = new Scanner(System.in);
        String input = scanner.nextLine();
        
        char[] characters = chars(input);
    }
    
    public static char[] chars(String arr) {
        ArrayList<Character> character = new ArrayList<Character>();

        for(int i = 0; i<arr.length();i++) {
            character.add(arr.charAt(i));           

            for(int k = 1; k<arr.length();k++) {
                if(character.get(i).equals(character.get(k))) {
                    character.remove(arr.charAt(i));                
                }
            }
        }
        System.out.print(character);
        return null;
    }
}

import java.util.*;

public class NoRepeats {
    
    public static void main(String args[]) {
        System.out.print("Please enter a phrase: ");
        Scanner scanner = new Scanner(System.in);
        String input = scanner.nextLine();
        //receive input
        char[] characters = chars(input);
        //call function
    }
    
    public static char[] chars(String arr) {
        ArrayList<Character> character = new ArrayList<Character>();
        //create an empty arraylist
        char[] characters = arr.toCharArray();
        
        for(int j = 0; j<characters.length;j++) {
            character.add(characters[j]);
        }
       
        for(int i = 0; i<characters.length;i++) {
            /*traverse through the orignal array adding a character to 
            the empty array list each time*/            
                for(int k = 0;k < characters.length;k++) {
                    if(characters[i]==characters[k]) {
                    System.out.print(character);                        
                    character.remove(characters[i]);                    
                }                   
            }
        }
        return null;
    }
}

预期投入:

“And I think to myself: what a wonderful world!”

预期产出:

And I thk o myself: w ru !

第一个错误是:

Exception in thread "main" java.lang.IndexOutOfBoundsException: Index 1 out of bounds for length 1 at java.base/jdk.internal.util.Preconditions.outOfBounds(Preconditions.java:64) at java.base/jdk.internal.util.Preconditions.outOfBoundsCheckIndex(Preconditions.java:70) at java.base/jdk.internal.util.Preconditions.checkIndex(Preconditions.java:266) at java.base/java.util.Objects.checkIndex(Objects.java:359) at java.base/java.util.ArrayList.get(ArrayList.java:427) at NoRepeats.chars(NoRepeats.java:20) at NoRepeats.main(NoRepeats.java:10)

第二种错误是:

"Exception in thread "main" java.lang.IndexOutOfBoundsException: Index 118 out of bounds for length 2 at java.base/jdk.internal.util.Preconditions.outOfBounds(Preconditions.java:64) at java.base/jdk.internal.util.Preconditions.outOfBoundsCheckIndex(Preconditions.java:70) at java.base/jdk.internal.util.Preconditions.checkIndex(Preconditions.java:266) at java.base/java.util.Objects.checkIndex(Objects.java:359) at java.base/java.util.ArrayList.remove(ArrayList.java:504) at NoRepeats.chars(NoRepeats.java:33) at NoRepeats.main(NoRepeats.java:11) "

Answer 1

First need to formulate your algorithm in words (or kind od pseudocode). Something like this: Problem: remove duplicated chars from string except space. Solution (pseudocode):

output_string = ""
used_chars = ""
for (c : input_string) {
  if (c ==    ) output_string += c;
  else if (!used_chars.contains(c)) {
    output_string += c;
    used_chars += c;
  }
}
print(output_string)

Now you can implement this in any language you want. Implementation in java can be:

static String solution(String input_string) {
    String output_string = "";
    Set<Character> used_chars = new TreeSet<>();
    for (char c : input_string.toCharArray()) {
        if (c ==    ) output_string += c;
        else if (!used_chars.contains(c)) {
            output_string += c;
            used_chars.add(c);
        }
    }
    return output_string;
}

public static void main(String[] args) {
    System.out.println(solution("And I think to myself: what a wonderful world!"));
}

Answer 2

<><>Code>:

package app;

import java.util.Scanner;

public class NoRepeats {
    public NoRepeats(){}

    public String scan(){
        System.out.print("Please enter a phrase: ");
        
        Scanner scanner = new Scanner(System.in);
        String read = scanner.nextLine(); //receive input
        scanner.close(); //do not forget to close
        
        return read;
    }

    public int contains(String character, char c, boolean caseInsensitive){
        int N = character.length();
        
        if(caseInsensitive) c = Character.toLowerCase(c);

        for(int i = 0; i < N; i++){
            char next = character.charAt(i);
            if(caseInsensitive) next = Character.toLowerCase(next);
            
            if(c == next) return i;
        }

        return -1;
    }

    public String getChars(String arr) {
        String character = "";

        for(int i = 0; i < arr.length(); i++){
            char next = arr.charAt(i);

            if(Character.isWhitespace(next) || contains(character, next, true) == -1) character += next;
            //else; //ignore
        }

        return character;
    }

    public static void main(String args[]) {
        NoRepeats noRepeats = new NoRepeats();

        String input = noRepeats.scan(),
               chars = noRepeats.getChars(input);

        System.out.println("String: " + chars);
    }
}

<>>Input:

And I think to myself: what a wonderful world!

<<>Output>:

String: And I thk o myself: w  ru !

Answer 3

你用<代码>i将特性阵列重新编号,在试图查阅内容时该阵列太短。

我建议你使用。

initialize a len to 0 to count the "good" characters.

allocate an array of the size of the original string to hold the result

if the character is a letter.

then add it to the set. If that returns false, it has been seen before so continue to the end of the for loop.

otherwise, add the character to the result array it is the first time

when done, copy the the len characters to an array and return it. See Arrays.copyOf for more info.

public static char[] chars(String arr) {
    Set<Character> seen = new HashSet<>();
    char[] result = new char[arr.length()];
    int len = 0;
    for (char ch : arr.toCharArray()) {
        if (Character.isLetter(ch))
            if (!seen.add(Character.toLowerCase(ch))) {
                continue; //skip to end of loop
            }
        
        result[len++] = ch;
    }
    return Arrays.copyOf(result, len);
    
}

这里是它如何工作。

String input = "And I think to myself: what a wonderful world!";
char[] result = chars(input);
System.out.println(new String(result));

印刷

And I thk o myself: w  ru !

友情链接