CODEWITHSUNDEEP
javaarrayshashhashtable
justinbg10
justinbg10

Reputation: 51

inserting random numbers into array Java

I have to hash 900 random integers into an empty table that has a set size of 1009 using open addressing. To determine where the number should go in the table I take the random number mod 1009 and then place the number there if it is open. If not I should check the next key after that and keep checking one by one until I find an open one to place the random number. The code I have so far is this: 

import java.util.*;

public class openAdd{
public static void main(String[] args) {
    //set table length
    int[] table = new int[1009];

    //insert 900 random integers into the table using open addressing
    //random number % table size = the key the number should be placed
    //if the key is already taken go to the next key until you find an open one
    Random randomGenerator = new Random();

    for (int i = 0; i < 900; i++) {
        int num = randomGenerator.nextInt(99999);
        int key = num % 1009;
        if (table[key] == 0) {
            table[key] = num;
        }
    }
}

}

I think what I have so far is good I am just confused on how to go about setting the key to key + 1 if there is already something in the original key. Thank you for your help, and just let me know if I need to add anything.

Upvotes: 2

Views: 173

Answers (1)

Jacob G.
Jacob G.

Reputation: 29680

You seem to have the right idea, just not the right implementation. If table[key] is nonzero, then you need to increment key until you find an index in table where table[key] is zero. You can utilize Java's remainder operator (like you already are) to prevent key from increasing over the bound of the array:

int key = num % 1009;

if (table[key] == 0) {
    table[key] = num;
} else {
    while (table[key = (key + 1) % table.length] != 0);
    table[key] = num;
}

Because table.length is greater than the amount of elements you're setting, there's no need to check if the array is full. Also, keep in mind that num can be 0.

Upvotes: 1

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