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->9->null, 14->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 table: [null, 9, null, null, null, 21, 14, null]

Given the original hash table, return the new hash table after rehashing .
Note
For negative integer in hash table, the position can be calculated as follow:

In 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.

In 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]

这道题就是根据条件老老实实的做

 1 /**
 2  * Definition for ListNode
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     /**
14      * @param hashTable: A list of The first node of linked list
15      * @return: A list of The first node of linked list which have twice size
16      */    
17     public ListNode[] rehashing(ListNode[] hashTable) {
18         // write your code here
19         int oldSize = hashTable.length;
20         int newSize = oldSize * 2;
21         if (hashTable==null || oldSize==0) return null;
22         ListNode[] res = new ListNode[newSize];
23         for (int i=0; i<oldSize; i++) {
24             if (hashTable[i] != null) rehash(hashTable, res, i);
25         }
26         return res;
27     }
28     
29     public void rehash(ListNode[] hashTable, ListNode[] res, int i) {
30         int newSize = res.length;
31         ListNode cur = hashTable[i];
32         while (cur != null) {
33             int val = cur.val;
34             int newPos = val>=0? val%newSize : (val%newSize+newSize)%newSize;
35             if (res[newPos] == null) res[newPos] = new ListNode(val);
36             else {
37                 ListNode temp = res[newPos];
38                 while (temp.next != null) {
39                     temp = temp.next;
40                 }
41                 temp.next = new ListNode(val);
42             }
43             cur = cur.next;
44         }
45     }
46 };
原文地址:https://www.cnblogs.com/EdwardLiu/p/4386374.html