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.
分析
旧表每条链表traverse,都hashcode一个新index进入新表(包括header!),如果新表位置已有node,就走到该node最后插入。
记得Node是链表,所以每次要new新的,不能直接连上旧的
/**
* 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 n = hashTable.length;
if(n == 0)
return hashTable;
int len = 2 * n;
ListNode[] nTable = new ListNode[len];
for(ListNode cur : hashTable){
while(cur != null){
int nIndex = hashcode(cur.val, len);
if(nTable[nIndex] == null){
nTable[nIndex] = new ListNode(cur.val);
}else{
ListNode head = nTable[nIndex];
while(head.next != null){
head = head.next;
}
head.next = new ListNode(cur.val);
}
cur = cur.next;
}
}
return nTable;
}
int hashcode(int key, int capacity) {
return (key % capacity + capacity) % capacity;
}
};
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