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Copy pathRehashing.java
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126 lines (97 loc) · 3.17 KB
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/*
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 .
Example
Given [null, 21->9->null, 14->null, null],
return [null, 9->null, null, null, null, 21->null, 14->null, null]
Note
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.
Tags Expand
LintCode Copyright Hash Table
Thoughts:
1. Loop through the hashtable[] and find longest, calcualte new capacity
2. rehash
*/
/**
* 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) {
if (hashTable == null || hashTable.length == 0) {
return hashTable;
}
//Find longest size
/*
int longest = 0;
for (int i = 0; i < hashTable.length; i++) {
ListNode node = hashTable[i];
int count = 0;
while (node != null) {
count++;
node = node.next;
}
longest = Math.max(longest, count);
}*/
//Calculate new capacity
//Just to clarify, this problem asks to double the hashtable size, rather than 'longest' times longer.
int capacity = hashTable.length * 2;
if (capacity == hashTable.length) {
return hashTable;
}
ListNode[] rst = new ListNode[capacity];
for (int i = 0; i < hashTable.length; i++) {
ListNode node = hashTable[i];
while (node != null) {
ListNode newNode = new ListNode(node.val);
int hCode = hashcode(newNode.val, capacity);
if (rst[hCode] == null) {
rst[hCode] = newNode;
} else {
ListNode move = rst[hCode];
while (move.next != null) {
move = move.next;
}
move.next = newNode;
}
node = node.next;
}
}
return rst;
}
public int hashcode(int key, int capacity) {
if (key < 0) {
return (key % capacity + capacity) % capacity;
} else {
return key % capacity;
}
}
};