Question

我有一个方法可以打印出非空的String数组元素。期望的输出：

1. cook
2. chef
3. baker
4. butcher
5. distiller

输出获取：

1. cook
3. chef
4. baker
7. butcher
9. distiller

这些数字并不像第一个例子中那样连续。显然这是因为当'i'不为空时它只打印'i'。无论如何我可以让它看起来像第一个例子？我尝试了不同的解决方案，但似乎都没有。

public class Main {

public void testMethod() {
    String myArray[] = new String [] { "cook", null, "chef", "baker", null, null, "butcher", null, "distiller" };
    for (int i = 0; i < myArray.length; i++) {
        if (myArray[i] != null) 
            System.out.println((i + 1) + ". " + myArray[i]);
    }   
}

public static void main(String[] args) {
    Main main = new Main();
    main.testMethod();
}
}

Answer 1

只需保留一个柜台：

public void testMethod() {
    String myArray[] = new String [] { "cook", null, "chef", "baker", null, null, "butcher", null, "distiller" };
    int j = 0;
    for (int i = 0; i < myArray.length; i++) {
        if (myArray[i] != null)  {
            System.out.println(++j + ". " + myArray[i]);
        }
    }

}

Answer 2

由于您无法使用i作为计数器，因此您需要一个单独的计数器：

public void testMethod() {
    String myArray[] = new String [] { "cook", null, "chef", "baker", null, null, "butcher", null, "distiller" };
    int counter = 1;
    for (int i = 0; i < myArray.length; i++) {
        if (myArray[i] != null) 
            System.out.println((counter++) + ". " + myArray[i]);
    }   
}

Answer 3

是的，只需记下您打印的数量，而不仅仅是指数：

- (void)willPresentSearchController:(UISearchController *)searchController {

    searchController.searchBar.tintColor = [UIColor whiteColor];
}

Answer 4

您需要一个新的整数变量来进行计数。试试这个 -

public class Main {
    public void testMethod() {
        int iter = 1;
        String myArray[] = new String [] { "cook", null, "chef", "baker", null, null, "butcher", null, "distiller" };
        for (int i = 0; i < myArray.length; i++) {
            if (myArray[i] != null) { 
                System.out.println((iter++) + ". " + myArray[i]);
            }
        }   
    }

    public static void main(String[] args) {
        Main main = new Main();
        main.testMethod();
    }
}

Answer 5

为count

再添加一个non-null values变量

public void testMethod() {
        String myArray[] = new String[] { "cook", null, "chef", "baker", null,
                null, "butcher", null, "distiller" };
        int j=0;
        for (int i = 0; i < myArray.length; i++) {
            if (myArray[i] != null){
                System.out.println((j + 1) + ". " + myArray[i]);
                j++;
            }
        }
    }

<强>输出

1. cook
2. chef
3. baker
4. butcher
5. distiller

Answer 6

请试一试。

public class Main {

    public void testMethod() {
        String myArray[] = new String [] { "cook", null, "chef", "baker", null, null, "butcher", null, "distiller" };

        int count = 0;

        for (int i = 0; i < myArray.length; i++) {
            if (myArray[i] != null) 
                System.out.println(((count++) + 1) + ". " + myArray[i]);
        }   
    }

    public static void main(String[] args) {
        Main main = new Main();
        main.testMethod();
    }
}

打印每个字符串

6 个答案: