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# Rehashing

## 题目描述

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 .

## 代码

public class ListNode {
int val;
ListNode next;
ListNode(int x) {
this.val = x;
next = null;
}
}
public class Solution {
public ListNode[] rehashing(ListNode[] hashTable) {
if (hashTable.length <= 0) return hashTable;
int newcapacity = 2 * hashtable.length;
ListNode[] newTable = new ListNode[newcapacity];
for (int i = 0; i < hashTable.length; i++) {
ListNode node = hashTable[i];
while (node != null) {
int newIndex = (node.val % newcapacity + newcapacity) % newcapacity;
if (newTable[newIndex] == null) {
newTable[newIndex] = new ListNode(node.val);
} else {
ListNode dummy = newTable[newIndex];
while (dummy.next != null) {
dummy = dummy.next;
}
dummy.next = new ListNode(node.val);
}
node = node.next;
}
}
return newTable;
}
}