Rehashing

Description

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]

Link

Lintcode_ladder

Method

1. x
2. x

Example

1. 1

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @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
     */ 
    vector<ListNode*> rehashing(vector<ListNode*> hashTable) {
        // write your code here
        if (hashTable.empty()) {
            return {};
        }
        int n = hashTable.size();
        int newsize = n*2;
        // consider that the new hashtable may just expend the size
        // So, the element's position(hash value) may be is the same as before
        // Input,size 5,      [null,null,42->32->null,-7->null,null]
        // Expected, size 10, [null,null,42->32->null,-7->null,null,null,null
        //                     null,null,null]
        vector<ListNode*> newhash(newsize, nullptr);
        for (int i = 0; i < n; ++i) {
            while (hashTable[i]) {
                int val = hashTable[i]->val;
                int pos = (val % newsize + newsize) % newsize;
                // insert to new pos
                ListNode* newcur = newhash[pos];
                if (!newhash[pos]) {
                    newhash[pos] = new ListNode(val);
                } else {
                    while (newcur->next) {
                        newcur = newcur->next;
                    }
                    newcur->next = new ListNode(val);
                }
                hashTable[i] = hashTable[i]->next;
            }
        }
        return newhash;
    }
};

Similar problems

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