在Java中按字母顺序对字符串进行排序而没有数组 [英] Sorting strings alphabetically without arrays in java

问题描述

因此，对于此程序，我需要从用户处获取三个字符串，然后按字母顺序对这些字符串进行排序.需要注意的是，我不能使用数组.同样，如果字符串不是以字母开头的，则不接受.到目前为止，我已经设法从用户​​那里获得了三个字符串，并且能够确定是否应该接受它们.我不知道如何按字母顺序对它们进行排序.我遇到的主要问题是忽略排序时不接受的字符串.这是从用户那里获取字符串的代码.

So for this program I need to obtain three strings from a user and sort those strings alphabetically. The caveat is that I cannot use arrays. Also if the string does not start with an alphabetic letter than it is not accepted. So far I have managed to get the three strings from the user and am able to determine if they should be accepted on or not. What I can't figure out is how to sort them alphabetically. The main issue I am having is ignoring the strings that weren't accepted when sorting. Here is the code to get the strings from the user.

    Scanner scan = new Scanner(System.in);

    System.out.print("Please input the first name: ");
    String name1 = scan.nextLine();
    name1 = name1.substring(0, 1).toUpperCase() + name1.substring(1,name1.length()).toLowerCase();
    if (name1.charAt(0) >= 'A' && name1.charAt(0) <= 'Z' ){
        System.out.println("The first name is: " + name1);
    }else{
        System.out.println("Error: The name was not accepted");
        name1 = null;
    }

    System.out.print("Please input the second name: ");
    String name2 = scan.nextLine();
    name2 = name2.substring(0, 1).toUpperCase() + name2.substring(1,name2.length()).toLowerCase();
    if (name2.charAt(0) >= 'A' && name2.charAt(0) <= 'Z'){
        System.out.println("The second name is: " + name2);
    }else{
        System.out.println("Error: The name was not accepted");
        name2 = null;
    }

    System.out.print("Please input the third name: ");
    String name3 = scan.nextLine();
    name3 = name3.substring(0, 1).toUpperCase() + name3.substring(1,name3.length()).toLowerCase();
    if (name3.charAt(0) >= 'A' && name3.charAt(0) <= 'Z'){
        System.out.println("The third name is: " + name3);
    }else{
        System.out.println("Error: The name was not accepted");
        name3 = null;
    }

推荐答案

我想出了这个解决方案.在不使用数组的情况下编写此代码是可读性的噩梦.

I came up with this solution. It is a nightmare in readability to write this code without an array.

import java.util.Scanner;

public class NameSort {

  private static final Scanner scan = new Scanner(System.in);

  public static void nameStort() {
    String name1 = readName("first");
    String name2 = readName("second");
    String name3 = readName("third");

    String potentialName1 = null;
    String potentialName2 = null;
    String potentialName3 = null;
    String potentialName4 = null;
    String potentialName5 = null;
    String potentialName6 = null;
    String potentialName7 = null;

    if (name1 == null) {
      if (name2 == null) {
        potentialName4 = name3;
      } else {
        potentialName4 = name2;
        if (name3 != null) {
          if (name2.compareTo(name3) > 0) {
            potentialName2 = name3;
          } else {
            potentialName6 = name3;
          }
        }
      }
    } else {
      potentialName4 = name1;
      if (name2 == null) {
        if (name3 != null) {
          if (name1.compareTo(name3) > 0) {
            potentialName2 = name3;
          } else {
            potentialName6 = name3;
          }
        }
      } else {
        if (name1.compareTo(name2) > 0) {
          potentialName2 = name2;
        } else {
          potentialName6 = name2;
        }
        if (name3 != null) {
          if (name1.compareTo(name3) > 0) {
            if (name2.compareTo(name3) > 0) {
              potentialName1 = name3;
            } else {
              potentialName3 = name3;
            }
          } else {
            if (name2.compareTo(name3) > 0) {
              potentialName5 = name3;
            } else {
              potentialName7 = name3;
            }
          }
        }
      }
    }
    System.out.println("The sorted names are: ");
    printIfNotNull(potentialName1);
    printIfNotNull(potentialName2);
    printIfNotNull(potentialName3);
    printIfNotNull(potentialName4);
    printIfNotNull(potentialName5);
    printIfNotNull(potentialName6);
    printIfNotNull(potentialName7);
  }

  private static void printIfNotNull(String potentialName) {
    if (potentialName != null) {
      System.out.println(potentialName);
    }
  }

  public static String readName(String position) {
    System.out.print("Please input the " + position + " name: ");
    String name = toTitleCase(scan.nextLine().trim());
    if (!name.isEmpty() && Character.isLetter(name.charAt(0))) {
      System.out.println("The " + position + " name is: " + name);
      return name;
    } else {
      System.out.println("Error: The name was not accepted");
    }
    return null;
  }

  public static String toTitleCase(String word) {
    if (word.isEmpty()) {
      return word;
    }
    // one way to do it
    return word.substring(0, 1).toUpperCase() + word.substring(1).toLowerCase();
  }

  public static void main(String[] args) {
    nameStort();
  }
}

但是在赋值中没有愚蠢的数组或集合限制的情况下，代码更具可读性:

But without the silly array or collection restriction in the assignment the code is much more readable:

import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class NameSort {

  private static final Scanner scan = new Scanner(System.in);

  public static void nameStort() {
    List<String> names = Arrays.asList(readName("first"), readName("second"), readName("third"));
    Collections.sort(names);
    for (String name : names) {
      printIfNotEmpty(name);
    }
  }

  private static void printIfNotEmpty(String name) {
    if (!name.isEmpty()) {
      System.out.println(name);
    }
  }

  public static String readName(String position) {
    System.out.print("Please input the " + position + " name: ");
    String name = toTitleCase(scan.nextLine().trim());
    if (!name.isEmpty() && Character.isLetter(name.charAt(0))) {
      System.out.println("The " + position + " name is: " + name);
      return name;
    } else {
      System.out.println("Error: The name was not accepted");
    }
    return "";
  }

  public static String toTitleCase(String word) {
    if (word.isEmpty()) {
      return word;
    }
    // one way to do it
    return word.substring(0, 1).toUpperCase() + word.substring(1).toLowerCase();
  }

  public static void main(String[] args) {
    nameStort();
  }
}

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