Rehashing

Med 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 .

Code

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) {
        int capacity = hashTable.length * 2;
        ListNode[] newTable = new ListNode[capacity];
        for (int i = 0; i < hashTable.length; i++) {
            ListNode node = hashTable[i];
            while(node != null) {
                int index = hash(node.val, capacity);
                ListNode rnode = newTable[index];
                if (rnode == null) {
                    newTable[index] = new ListNode(node.val);
                } else {
                    ListNode newNode = new ListNode(node.val);
                    while (rnode.next != null) {
                        rnode = rnode.next;
                    }
                    rnode.next = newNode;
                }
                node = node.next;
            }
        }

        return newTable;
    }

    private int hash(int key, int capacity) {
        return (key % capacity + capacity) % capacity;
    }
};

Analysis

最坏$O(n^2)$