Rehashing

Medium Rehashing

25%

Accepted

The size of the hash table is not determinate at the very beginning. If the total size of keys is too large (e.g. size >= capacity / 10), we should double the size of the hash table and rehash every keys. Say you have a hash table looks like below:

size=3, capacity=4

[null, 21, 14, null]
       ↓    ↓
       9   null
       ↓
      null

The hash function is:

int hashcode(int key, int capacity) {
    return key % capacity;
}

here we have three numbers, 9, 14 and 21, where 21 and 9 share the same position as they all have the same hashcode 1 (21 % 4 = 9 % 4 = 1). We store them in the hash table by linked list.

rehashing this hash table, double the capacity, you will get:

size=3, capacity=8

index:   0    1    2    3     4    5    6   7
hash : [null, 9, null, null, null, 21, 14, null]

Given the original hash table, return the new hash table after rehashing .

Have you met this question in a real interview? 

Yes

Example

Given [null, 21->9->null, 14->null, null],

return [null, 9->null, null, null, null, 21->null, 14->null, null]

Note

For negative integer in hash table, the position can be calculated as follow:

• C++/Java: if you directly calculate -4 % 3 you will get -1. You can use function: a % b = (a % b + b) % b to make it is a non negative integer.
• Python: you can directly use -1 % 3, you will get 2 automatically.

/**

 * Definition for ListNode

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) {

 *         val = x;

 *         next = null;

 *     }

 * }

 */

public class Solution {

    /**

     * @param hashTable: A list of The first node of linked list

     * @return: A list of The first node of linked list which have twice size

     */    

    public ListNode[] rehashing(ListNode[] hashTable) {

        // write your code here

        int len = hashTable.length * 2;

        ListNode[] newTable = new ListNode[len];

        

        for(int i = 0; i < hashTable.length; i++){

            

            ListNode node = hashTable[i];

            

            while(node != null){

                

                int index = node.val >= 0 ? node.val % len : (node.val % len + len) % len;

                if(newTable[index] == null){

                    newTable[index] = new ListNode(node.val);

                } else{

                    ListNode temp = newTable[index];

                    while(temp.next != null){

                        temp = temp.next;

                    }

                    temp.next = new ListNode(node.val);

                }

                node = node.next;

            }

        }

        return newTable;

        

        

    }

};