Question

我知道已经有一些问题在问与我的问题类似的东西，但是尽管阅读了这些内容，但是他们并没有完全按照我的意愿去做。我正在创建一个代码，接受用户输入的0-100（含）之间的数字。无论数字是多少，它都会打印出导致该数字和该数字产生的所有数字

EX：用户输入= 25 输出= 012345678910111213141516171819202122232425

我正在努力。现在，我应该使用该字符串并创建两个新字符串，一个仅用于奇数，另一个用于偶数。

EX：用户输入= 25 输出：奇数：135791113151719212325＆偶数= 024681012141618202224

到目前为止，这是我的代码：

import java.util.Scanner;

public class OddAndEven{
public String quantityToString() {
  Scanner number = new Scanner(System.in);
  int n = number.nextInt();
  String allNums = "";
  if ((n >= 0) && (n <= 100)) {
     for (int i = 0;i <= n; ++i)
     allNums = allNums + i;

     return allNums;
  }
  else {
     return "";
  }


}

  public void oddAndEvenNumbers(int num) {//Start of second method

     String allNums = ""; //String that quantityToString returns
     String odd = "";
     String even = "";

     if ((num >= 0) && (num < 10)) { //Looks at only single digit numbers
        for (int i = 0; i <= allNums.length(); i++) {
           if (Integer.parseInt(allNums.charAt(i))%2 == 0) { //trying to get the allNums string to be broken into individual numbers to evaluate
           even = even + allNums.charAt(i); //adding the even numbers of the string
           }
           else {
              odd = odd + allNums.charAt(i);
           }
        }
     }
     else { //supposed to handle numbers with double digits
        for (int i = 10; i <= allNums.length(); i = i + 2) {
           if (Integer.parseInt(allNums.charAt(i))%2 == 0) {
              even = even + allNums.charAt(i);
           }
           else {
              odd = odd + allNums.charAt(i);
           }
        }
     }
  System.out.println("Odd Numbers: " + odd);
  System.out.println("Even Numbers: " + even);

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

System.out.println(new OddAndEven().quantityToString());
//System.out.println(new OddAndEven().oddAndEvenNumbers(allNums));
   //Testing     

  OddAndEven obj = new OddAndEven();
  System.out.println("Testing n = 5");
  obj.oddAndEvenNumbers(5);
  System.out.println("Testing n = 99");
  obj.oddAndEvenNumbers(99);

我知道我的问题在于应该将字符串拆开并评估各个数字，但是我不知道该怎么办。（我也尝试过substring（）和trim（））而且我还没有学会如何使用数组，所以这就是为什么我没有尝试使用数组的原因。

Answer 1

我认为您可以那样做：

    int x = 20;
    StringBuilder evenNumberStringBuilder = new StringBuilder();
    StringBuilder oddNumberStringBuilder = new StringBuilder();
    for(int i =0 ; i<x+1; i++){
        if(i % 2 == 0)evenNumberStringBuilder.append(i);
        else oddNumberStringBuilder.append(i);
    }

    System.out.println(evenNumberStringBuilder);
    System.out.println(oddNumberStringBuilder);

输出：

02468101214161820
135791113151719

Answer 2

您已经将输入作为整数，所以不要使用字符串。我建议使用此循环；

    Scanner number = new Scanner(System.in);

    System.out.print("Even Numbers: ");
    for (int i = 0; i <= number; i=i+2) {
     System.out.print(i);
    }

    System.out.println("");

    System.out.print("Odd Numbers: ");
    for (int i = 1; i <= number; i=i+2) {
     System.out.print(i);
    }

Answer 3

您可以在将数字存储在allnumbers字符串中时简单地评估数字，这是一个有效的代码：

<div class="form-row">
  <div class="form-group needs-validation col-sm-6 col-md-6">
    <label for="inputCity">City <small class="text-muted">(required)</small></label>
    <input type="text" class="form-control" id="inputCity" name="city" required>
    <div class="invalid-feedback">
      Please provide your city.
    </div>
  </div>
  <div class="form-group needs-validation col-sm-6 col-md-4">
    <label for="inputState">State <small class="text-muted">(required)</small></label>
    <input type="text" class="form-control" id="inputState" name="state" required
              autocomplete="address-level1">
    <div class="invalid-feedback">
      Please provide your state.
    </div>
  </div>
  <div class="form-group needs-validation col-5 col-sm-4 col-md-2">
    <label for="inputZip">Zip <small class="text-muted">(required)</small></label>
    <input type="text" class="form-control" id="inputZip" name="zip" required>
    <div class="invalid-feedback">
      Please provide your zip code.
    </div>
  </div>
</div>

只有偶数和只有奇数的字符串

3 个答案: