Rehashing

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 .

Notice

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.

public ListNode[] rehashing(ListNode[] hashTable) {
    if(hashTable == null || hashTable.length == 0) {
        return new ListNode[0];
    }
    int len = hashTable.length * 2;
    ListNode[] res = new ListNode[len];
    for(int i = 0; i < hashTable.length; i++) {
        ListNode p = hashTable[i];
        while(p != null) {
            if(res[(p.val % len + len) % len] == null) {
                ListNode node = new ListNode(p.val);
                res[(p.val % len + len) % len] = node;
            }
            else {
                ListNode pn = res[(p.val % len + len) % len];
                while(pn.next != null) {
                    pn = pn.next;
                }
                pn.next = new ListNode(p.val);
            }
            p = p.next;
        }
    }
    return res;
}