Rehashing

Question

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 .

Thoughts

简单，loop through all the elements, 如果是linkedlist就一个一个处理过去

Solution

"""
Definition of ListNode
class ListNode(object):

    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
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
    """
    def rehashing(self, hashTable):
        # write your code here
        cap = len(hashTable)
        newCap = cap * 2
        newHash = [ None for i in range(newCap)]
        for l in hashTable:
            while l:
                newEle = ListNode(l.val)
                newPos = l.val % newCap
                newListHead = newHash[newPos]
                if not newListHead:
                    newHash[newPos] = newEle
                else:
                    while newListHead.next:
                        newListHead = newListHead.next
                    newListHead.next = newEle
                l = l.next

        return newHash