Get Index number of an element for all occurences-CodePudding

I would like to get the index number of an element, the following code gets the first occurrence only.

Array content for hair:

short,
long,
long

What index number I get: 1

What index number is needed: 1, 2


            for(int i = 0; i <= hair.length; i=i 1) {
                if(hair[i].equalsIgnoreCase("long")) {
                    index = i;
                    break;
                }
           }

CodePudding user response：

You could do something like this

    public static void main(String[] args) {
        String[] hair = {"short", "long", "long"}; 
        int[] indexes = new int[hair.length];
        for(int i = 0; i < hair.length; i  ) {
            if(hair[i].equalsIgnoreCase("long")) {
                indexes[i] = i;
            }else{
                indexes[i] = -1;
            }
        }
           
       for(int i = 0; i < indexes.length; i  ) {
            if(indexes[i]!=-1){
                System.out.println(indexes[i]);
            }
       }
    }

Demo

CodePudding user response：

You may use a container such as a List to store each valid index, also careful the stop condition should be i < hair.length

String[] hair = {"short", "long", "long"};
List<Integer> result = new ArrayList<>();
for (int i = 0; i < hair.length; i  ) {
    if (hair[i].equalsIgnoreCase("long")) {
        result.add(i);
    }
}
System.out.println(result); // [1, 2]

Use a int[] to store is a bit different, as you'll have less valid index as the size of the array, ending with empty box at the end (0 value), you'll need a Arrays.copyof to truncate it

String[] hair = {"short", "long", "long", "short", "long", "long", "short", "long", "long"};
int[] indices = new int[hair.length];
int nb_valid = 0;
for (int i = 0; i < hair.length; i  ) {
    if (hair[i].equalsIgnoreCase("long")) {
        indices[nb_valid  ] = i;
    }
}
System.out.println(Arrays.toString(indices)); // [1, 2, 4, 5, 7, 8, 0, 0, 0]

indices = Arrays.copyOf(indices, nb_valid); // truncate the array to keep real indices only
System.out.println(Arrays.toString(indices)); // [1, 2, 4, 5, 7, 8]

CodePudding user response：

The "break;" inside the loop is breaking the whole "for" loop.

As said, don't override the "index" param (instead, use some datastructure to hold more than 1 object), and remove the "break", so you can iterate all elements of the "hair" array.

CodePudding user response：

You may use an int array to hold the index values.

Here's the input

{ "short", "long", "long", "short", "long" }

Here's the output

1 2 4

Here's the complete runnable code.

public class FindMatches {

    public static void main(String[] args) {
        String[] hair = { "short", "long", "long", "short", "long" };
        int[] indexValues = new int[hair.length];
        int index = 0;

        for (int i = 0; i < hair.length; i  ) {
            if (hair[i].equalsIgnoreCase("long")) {
                indexValues[index  ] = i;
            }
        }

        for (int i = 0; i < index; i  ) {
            System.out.print(indexValues[i]);
            if (i < (index - 1)) {
                System.out.print(" ");
            }
        }

        System.out.println();
    }

}

CodePudding user response：

You can also use streams for finding matching indices

String[] hair = {"short", "long", "long"};
int[] indices = IntStream.range(0, hair.length)
            .filter(i -> "long".equalsIgnoreCase(hair[i]))
            .toArray();