Question

我需要返回仅包含奇数的数组，例如[1,3,5]。在学校教育中，我被要求这样做，但我看不到哪里出了问题。

public static int[] odds(int numOdds) {
    int[] odds = numOdds;
    for (int i=0; i<odds.length; i++) {
        if (odds[i] %2 != 0) {
            return odds;
        }
    }
}

public static void main(String[] args) {
    int[] theOdds = odds(3);
    System.out.println(theOdds[0] + ", " + theOdds[1] + ", " + theOdds[2]);
}

Answer 1

如果要从数组中返回赔率，则应首先将其作为参数传递，也应将IF语句后的赔率存储在新数组中。在可以轻松地将ArrayList转换为普通Array之后，由于不知道有多少几率，因此应该在第一次使用动态列表。像这样：

public static int[] odds(int[] arrayOfNumber) {

List<Integer> odds = new ArrayList<Integer>();

for (int i=0; i<arrayOfNumber.length; i++) {
    if (arrayOfNumber[i] %2 != 0) {
        odds.add(arrayOfNumber[i]);
    }
  }
  return odds.toArray();
}

Answer 2

这是您发布的代码。在下面查看我的评论

public static int[] odds(int numOdds) {
    //   int[] odds = numOdds;     // needs to be int[] odds = new int[numOdds] to 
                                   //store the values.  It is the array you will 
                                   // return.
    int[] odds = new int[numOdds];

    int start = 1;            // you need to have a starting point
    for (int i=0; i<odds.length; i++) {
    //    if (odds[i] %2 != 0) { // don't need this as you are generating odd 
                                 // numbers yourself


    //    return odds;      // You're doing this too soon.  You need to store 
                               // the numbers in the array first 

          odds[i] = start;  // store the first odd number
          start += 2;       // this generates the next odd number.  Remember that
                            // every other number is even or odd depending from
                            // where you start.

    }
    return odds;         // now return the array of odd numbers.

    }
}

public static void main(String[] args) {
    int[] theOdds = odds(3);
    System.out.println(theOdds[0] + ", " + theOdds[1] + ", " + theOdds[2]);
}

Answer 3

您可以通过本机Java 8流API节省时间：

int[] odds = Arrays.stream(arrayOfNumbers).filter(number -> number%2 != 0).toArray();

Streams提供了许多方法来使您的工作迅速

返回只有奇数的数组

3 个答案: