Rehashing

update Aug 22, 2017

LintCodearrow-up-right

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

Basic Idea:

其实就是rehashing的基本操作,用linkedList解决collision;

Java Code: