Rehashing

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

/**
 * 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) {
        if (hashTable == null || hashTable.length == 0) {
            return hashTable;
        }
        int capacity = hashTable.length;
        int newCapacity = 2 * capacity;
        ListNode[] newHashTable = new ListNode[newCapacity];
        for (int i = 0; i < capacity; i++) {
            ListNode ln = hashTable[i];
            while (ln != null) {
                int code = hashcode(ln.val, newCapacity);
                insertToHashTable(newHashTable, code, ln.val);
                ln = ln.next;
            }
        }
        return newHashTable;
    }
    public int hashcode(int key, int capacity) {
        int hash;
        if (key < 0) {
            hash = (key % capacity + capacity) % capacity;
        } else {
            hash = key % capacity;
        }
        return hash;
    }
    private void insertToHashTable(ListNode[] hashTable, int code, int value) {
        if (code < hashTable.length) {
            ListNode ln = hashTable[code];
            if (ln == null) {
                hashTable[code] = ln = new ListNode(value);
            } else {
                while (ln.next != null) {
                    ln = ln.next;
                }
                ln.next = new ListNode(value);
            }
        }
    }

    public static void main(String[] args) {
        Solution s = new Solution();
        ListNode[] lsn = new ListNode[3];
        lsn[0] = null;
        lsn[1] = null;
        lsn[2] = new ListNode(29);
        lsn[2].next = new ListNode(5);
        ListNode[] newLsn = s.rehashing(lsn);
    }
};

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
        next = null;
    }
}