[LintCode] Rehashing

Jason 发布于2019-08-14 15:25 / 723人阅读

Problem

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]

Note Solution

 public class Solution {
        public ListNode[] rehashing(ListNode[] hashTable) {
            if (hashTable == null || hashTable.length == 0) return hashTable;
            int cap = hashTable.length*2;
            ListNode[] res = new ListNode[cap];
            ListNode[] temp = new ListNode[cap];
            for (int i = 0; i < cap/2; i++) {
                ListNode cur = hashTable[i];
                while (cur != null) {
                    ListNode node = new ListNode(cur.val);
                    hashTable[i] = cur.next;
                    int index = (cur.val%cap + cap)%cap;
                    if (res[index] == null) {
                        res[index] = node;
                        temp[index] = node;
                    }
                    else {
                        temp[index].next = node;
                        temp[index] = node;
                    }
                    cur = cur.next;
                }
            }
            return res;
        }
    };

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