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   7hash : [null, 9, null, null, null, 21, 14, null]

Given the original hash table, return the new hash table after rehashing .

rehash意思是重hash,　即原来的key的数目太高，需要将hash table的大小扩大二倍，然后重新获得key值．

了解概念的题目．

代码如下：

"""Definition of ListNodeclass 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):        if not hashTable:            return []        capacity = len(hashTable)*2        rehash = [None]*capacity        #insert on the head of linked list        for i in xrange(len(hashTable)):            cur = hashTable[i]            while cur:                key = cur.val%capacity                if not rehash[key]:                    rehash[key] = ListNode(cur.val)                else:                    tmp = rehash[key]                    while tmp.next:                        tmp = tmp.next                    tmp.next = ListNode(cur.val)                cur = cur.next        return rehash