Question

我首先需要从用户处接收未知数量的字符串输入，并创建一个循环，该循环将在输入“ x”时终止我需要跟踪输入了多少输入我需要测试输入以获取最长和最短的字符串及其长度我需要找到字符串长度的总和我需要找到平均字符串长度我需要找到输入的字符串数量（不包括“ x”

）

Answer 1

首先让我强调一下，您应该真正尝试首先自己实现它，因为否则您将永远无法正确理解逻辑/语法，..

不存储值的操作可以如下：

我只是不确定平均长度是否可以正常工作，但看起来应该可以

（检查代码中的注释计算，这种方式不起作用，最后打印可能有效）

public static void getStringsAndGetDetails() {
    //https://stackoverflow.com/questions/54689213/i-would-like-to-find-out-how-to-find-the-shortest-and-longest-strings-from-a-lis
    System.out.println("Put 'x' or 'X' for exit");
    String input ="";
    Scanner sc = new Scanner(System.in);

    String shortestStr = null;
    String longestStr = null;

    int counter = 0;
    //int averageLength = 0;

    //load till x or X is entered 
    do{
        //read
        System.out.print("please, enter input: ");
        input = sc.nextLine();

        //avoid to process when escape is entered
        if(!input.toLowerCase().equals("x")) {
            counter++;

            //get to temp variable, can be reused with avoid repeating reading
            int inputLength = input.length();

            //if input is shorter then its new shortest, null check for first input
            if(shortestStr == null || inputLength < shortestStr.length()) {
                shortestStr = input;
            }

            //similar with longest string
            if(longestStr == null || inputLength > longestStr.length()) {
                longestStr = input;
            }

            //!! calculate average length is NOT possible like that, see bellow in printing
            /*
            averageLength = (averageLength + inputLength) / counter;
            System.out.println();
            */
        }
    }while(!input.toLowerCase().equals("x"));
    //not mandatory, but best practice (closing the streams), scanner in this case
    sc.close();


    System.out.println("Entered " + counter + " values");
    System.out.println("Longest: "+ longestStr);
    System.out.println("Shortest: "+ shortestStr);

    //be sure data are available (in case you enter x in first iteration, will fail otherwise)
    if(shortestStr != null && longestStr != null) {
    //seems to work
        int averageLength= (shortestStr.length() + longestStr.length()) / 2;  
        System.out.println("Average length: "+ averageLength);  
    }

}

输出：

Put 'x' or 'X' for exit
please, enter input: kakaka
please, enter input: ka
please, enter input: kaka
please, enter input: x
Entered 3 values
Longest: kakaka
Shortest: ka
Average length: 4

Answer 2

看看下面的代码。它包含一些您希望在程序中实现的良好实践。注意测试是如何在main方法中编写的吗？我从这些开始，所以我可以在编写实际方法之前仔细考虑解决方案。在测试中，我尝试捕获尽可能多的边缘情况。如果我在代码中发现错误，则可以编写一个新的单元测试以使其失败，然后修复代码。免责声明：此代码需要将JUnit 5添加到您的类路径中。

import static org.junit.Assert.assertEquals;
import static org.junit.Assert.assertTrue;

public class StringsInfo {

  public static void main(String[] args) {
    StringsInfo sInfo = new StringsInfo();

    sInfo.test(new String[] {}, null, null, 0, 0.0D, 0);
    sInfo.test(new String[] {null}, null, null, 0, 0.0D, 0);
    sInfo.test(new String[] {"one"}, "one", "one", 1, 3.0D, 3);
    sInfo.test(new String[] {"one", "two"}, "one", "one", 2, 3.0D, 6);
    sInfo.test(new String[] {"one", "two", "three"}, "one", "three", 3, (11/3), 11);
    sInfo.test(new String[] {"one", "two", "three", "x", "five"}, "one", "three", 3, (11/3), 11);
  }

  /**
   * Helper to test the class
   * @param input
   * @param shortest
   * @param longest
   * @param count
   * @param average
   * @param sumLength
   */
  private void test(String[] input, String shortest, String longest, long count,
      double average, long sumLength) {
    Result result = processString(input);
    assertEquals(shortest, result.shortest);
    assertEquals(longest, result.longest);
    assertEquals(count, result.count);
    assertTrue(average == result.average);
    assertEquals(sumLength, result.sumLength);
  }

  private Result processString(String[] input) {
    Result result = new Result();
    int shortestLength = Integer.MAX_VALUE;
    int longestLength = Integer.MIN_VALUE;
    for (String s : input) {
      if (s == null) { // Ignore null Strings
        continue;
      }
      if ("x".equals(s)) { // found the end
        break;
      }
      result.count++;
      int wordLength = s.length();
      result.sumLength += wordLength;
      if (wordLength < shortestLength) {
        result.shortest = s;
        shortestLength = wordLength;
      }
      if (wordLength > longestLength) {
        result.longest = s;
        longestLength = wordLength;
      }
    }
    if (result.count > 0) { // avoid division by zero
      result.average = result.sumLength / result.count;
    }
    return result;
  }

}

/**
 * Helper class to store the result
 */
class Result {
  String shortest;
  String longest;
  long count;
  double average;
  long sumLength;
}

如何从字符串输入列表中找到最短和最长的字符串

2 个答案: