L129 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.

Example

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

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

public ListNode[] rehashing(ListNode[] hashTable) {
    if (hashTable == null || hashTable.length == 0) {
        return new ListNode[2];
    }

    int size = hashTable.length;
    ListNode[] res = new ListNode[size * 2];
    for (ListNode lNode : hashTable) {
        while (lNode != null) {
            int val = lNode.val;
            add(val, res);
            lNode = lNode.next;
        }
    }
    return res;
}

private void add(int val, ListNode[] res) {
    int loc = -1;
    int size = res.length;
    if (val >= 0) {
        loc = val % size;
    } else {
        loc = (val % size + size) % size;
    }

    ListNode tmp = new ListNode(val);
    if (res[loc] == null) {
        res[loc] = tmp;
    } else {
        ListNode head = res[loc];
        while (head.next != null) {
            head = head.next;
        }
        head.next = tmp;
    }
}