[LintCode] 129 Rehashing 解题报告

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]



思路
建立一个比原来大一倍的表。
遍历原来表里的每一个元素，发配到新表相应的位置。



Code

/**
 * 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
        
        int capacity = hashTable.length * 2;
        ListNode[] newTable = new ListNode[capacity];
        
        for (ListNode head : hashTable) {
            while (head != null) {
                int newIndex = (capacity + head.val % capacity) % capacity;
                if (newTable[newIndex] == null) {
                    newTable[newIndex] = new ListNode(head.val);
                }
                else {
                    ListNode curr = newTable[newIndex];
                    while (curr.next != null) {
                        curr = curr.next;
                    }
                    curr.next = new ListNode(head.val);
                }
                head = head.next;
            }
        }
        
        return newTable;
    }
};