java - 数组值输出多个

Question

这是我想出的代码。但是，我希望能够输出:

• (值)位于槽 x 中。
• (值)位于槽 x 中。

如果(值)在两个槽中可用(例如 7)，则有两个输出。

或

• (num) 不在数组中。

但不能两者兼而有之。有人可以帮忙吗？

    public static void main(String[] args) {
    int search, counter;
    int num[]={3, 4, 5, 6, 7, 8, 10, 7,  9, 13};

    System.out.print("Array: ");
    for (int count=0; count<num.length; count++)
        System.out.print(+num[count] + " ");

    Scanner in = new Scanner (System.in);
    System.out.print("\nValue to find: ");
    search = in.nextInt();

    for (counter = 0; counter < num.length; counter++ ){
        if (num[counter] == search)
        {
            System.out.println(search + " is in slot " + (counter + 1) + ".");
        }           
    }
    if (counter == num.length)
        {
            System.out.println(search + " is not in the array.");
        }
}
}

Answer 1

虽然我觉得你应该在另一个社区上问这样的问题，例如 https://codereview.stackexchange.com/我可以提供一个建议:

使用 boolean 标志来检查您之前是否找到过它。像这样的事情:

public static void main(String[] args) {
  int search;
  boolean found = false;
  int num[]={3, 4, 5, 6, 7, 8, 10, 7,  9, 13};

  System.out.print("Array: ");
  for (int count=0; count<num.length; count++)
      System.out.print(+num[count] + " ");

  Scanner in = new Scanner (System.in);
  System.out.print("\nValue to find: ");
  search = in.nextInt();

  for (int counter = 0; counter < num.length; counter++ ) {
      if (num[counter] == search)
      {
        System.out.println(search + " is in slot " + (counter + 1) + ".");
        found = true;
      }           
  }

  if (!found) {
        System.out.println(search + " is not in the array.");
  }

  in.close();

}

因此，只有在线性遍历数组后找不到元素时，才会打印“未找到”消息...