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 .
Example:
Given [null, 21->9->null, 14->null, null],
return [null, 9->null, null, null, null, 21->null, 14->null, null]
Solution:
the hastable is passed in we can do double it direclty
进来后就直接double, 不用考虑是否要double的问题。
数组代替hash表,所以是数组操作。
"""
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
newHashTable = [None for i in range(len(hashTable) * 2)]
for item in hashTable:
node = item
while node:
val = node.val
self.addNewNode(newHashTable, val)
node = node.next
return newHashTable
def addNewNode(self, newHashTable, val):
key = val % len(newHashTable)
if newHashTable[key]:
self.addListNode(newHashTable[key], val)
else:
newHashTable[key] = ListNode(val)
def addListNode(self,node, val):
if node.next:
self.addListNode(node.next, val)
else:
node.next = ListNode(val)
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