Problem
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
Given [null, 21->9->null, 14->null, null],
return [null, 9->null, null, null, null, 21->null, 14->null, null]
Note
Solution
public class Solution {
public ListNode[] rehashing(ListNode[] hashTable) {
if (hashTable == null || hashTable.length == 0) return hashTable;
int cap = hashTable.length*2;
ListNode[] res = new ListNode[cap];
ListNode[] temp = new ListNode[cap];
for (int i = 0; i < cap/2; i++) {
ListNode cur = hashTable[i];
while (cur != null) {
ListNode node = new ListNode(cur.val);
hashTable[i] = cur.next;
int index = (cur.val%cap + cap)%cap;
if (res[index] == null) {
res[index] = node;
temp[index] = node;
}
else {
temp[index].next = node;
temp[index] = node;
}
cur = cur.next;
}
}
return res;
}
};