LintCode-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->9->null, 14->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 table: [null, 9, null, null, null, 21, 14, null]

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

Note

For negative integer in hash table, the position can be calculated as follow:

In
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.

In 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]

Solution:

 1 /**
 2  * Definition for ListNode
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     /**
14      * @param hashTable: A list of The first node of linked list
15      * @return: A list of The first node of linked list which have twice size
16      */
17     public ListNode[] rehashing(ListNode[] hashTable) {
18         int size = hashTable.length;
19         if (size==0) return null;
20         int newSize = 2*size;
21         ListNode[] newTable = new ListNode[newSize];
22         Arrays.fill(newTable,null);
23
24         for (int i=0;i<size;i++){
25             ListNode cur = hashTable[i];
26             while (cur!=null){
27                 ListNode temp = cur;
28                 cur = cur.next;
29                 //Calculate the new position for temp.
30                 int val = (temp.val % newSize + newSize) % newSize;
31                 if (newTable[val]==null){
32                     newTable[val]=temp;
33                     temp.next = null;
34                 } else {
35                     ListNode p = newTable[val];
36                     while (p.next!=null) p = p.next;
37                     p.next = temp;
38                     temp.next = null;
39                 }
40             }
41         }
42
43         return newTable;
44     }
45 };