Rehashing

时间:2023-03-08 22:07:20

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=3capacity=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=3capacity=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.
Have you met this question in a real interview?
Yes
Example

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

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

 /**
* Definition for ListNode
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public 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
*/
public ListNode[] rehashing(ListNode[] hashTable) {
if (hashTable == null || hashTable.length == ) return hashTable; ListNode[] newHT = new ListNode[hashTable.length * ]; for (int i = ; i < hashTable.length; i++) {
ListNode current = hashTable[i];
while (current != null) {
ListNode next = current.next;
current.next = null;
add(newHT, current);
current = next;
}
}
return newHT;
} public void add(ListNode[] hashTable, ListNode node) {
int index = node.val % hashTable.length;
if (index < ) {
index = hashTable.length + index;
}
if (hashTable[index] == null) {
hashTable[index] = node;
} else {
ListNode current = hashTable[index];
while (current.next != null) {
current = current.next;
}
current.next = node;
}
}
};