Rehashing(重哈希)

问题

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.
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]

分析

需要注意的是在新数组中赋值时需要重新new一个对象。其它的请参阅 HashMap去重原理和内部实现

代码

 * 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) {
        // write your code here
        ListNode[] res=new ListNode[hashTable.length*2];
        for(ListNode node:hashTable){
            while(node!=null){
                int index=(node.val%res.length+res.length)%res.length;
                ListNode temp=res[index];
                if(temp==null){
                    res[index]=new ListNode(node.val);
                }else{
                    while(true){
                        if(temp.next==null){
                            temp.next=new ListNode(node.val);
                            break;
                        }else{
                            temp=temp.next;
                        }
                    }
                }
                node=node.next;
            }
        }
        return res;
    }
};