[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, 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 .

Example

Given [null, 21->9->null, 14->null, null],

return [null, 9->null, null, null, null, 21->null, 14->null, null]

Note

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.

/**
 * 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
     */    
    int getLen(ListNode *n)
    {
        int len = 0;
        while(n)
        {
            n=n->next;
            len++;
        }
        return len;
    }
    vector<ListNode*> rehashing(vector<ListNode*> hashTable) {
        // write your code here
        int nums = 0;
        int len = hashTable.size();
        if(0 == len) return hashTable;
        
        for(auto e:hashTable)
        {
            nums += getLen(e);
        }
        if(nums < len/10) return hashTable;
        
        int cap = 2*len;
        vector<ListNode*> res(cap, NULL);
        
        for(auto e:hashTable)
        {
            if(!e) continue;
            while(e)
            {
                ListNode *p = e->next;
                int index = (e->val%cap + cap)%cap;
                if(res[index] == NULL) res[index] = e;
                else {
                    ListNode *t = res[index];
                    while(t->next) t = t->next;
                    t->next = e;
                }
                e->next = NULL;
                e = p;
            }
        }
        return res;
    }
};