Question

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]

Analysis

此题的难度不大，只需要按照题目的要求实现代码就可以。不过需要注意的是：

1. C++/Java中，不能直接对负数使用取模运算，而需要用等式 a % b = (a % b + b) % b ，让所得到的hash值为非负数。
2. 所得到的新的HashTable中，可能依然存在碰撞，所以仍然需要在对应hashcode位置的ListNode tail上插入新的ListNode。

Solution

主要是关注老table和新table在一个index上如果有多个node需要loop

老的nodes：while(hashTable[i] != null)

新的nodes： 需要dummy>>> loop到最后dummy.next ==null; 这个时候加入新的node

/**
 * 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
        if(hashTable == null ||hashTable.length == 0 ){
            return hashTable;
        }

        int newCapacity = 2*hashTable.length;

        ListNode[] newTable = new ListNode[newCapacity];
        for(int i = 0; i < hashTable.length; i++){
            while(hashTable[i] != null){
                int newIndex = 
                (hashTable[i].val%newCapacity + newCapacity)%newCapacity;

                if(newTable[newIndex] == null){
                   newTable[newIndex] = new ListNode(hashTable[i].val); 
                }else{
                    ListNode dummy = newTable[newIndex];
                    while(dummy.next != null){
                        dummy = dummy.next;
                    }
                    dummy.next = new ListNode(hashTable[i].val); 
                }
                hashTable[i] = hashTable[i].next;        
            }
        }
            return newTable;
    }
};