lintcode-medium-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 .

 

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]

 

/**
 * 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 size = hashTable.length;
        ListNode[] res = new ListNode[2 * size];
        
        for(int i = 0; i < size; i++){
            if(hashTable[i] != null){
                ListNode p = hashTable[i];
                
                while(p != null){
                    int index = (p.val % (2 * size) + 2 * size) % (2 * size);
                    if(res[index] == null){
                        res[index] = new ListNode(p.val);
                    }
                    else{
                        ListNode temp = res[index];
                        while(temp.next != null){
                            temp = temp.next;
                        }
                        temp.next = new ListNode(p.val);
                    }
                    
                    p = p.next;
                }
            }
        }
        
        return res;
    }
};

 

转载于:https://www.cnblogs.com/goblinengineer/p/5352235.html