LintCode 129. Rehashing 原创Java参考解答
问题描述
http://www.lintcode.com/en/problem/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.
Example
Given [null, 21->9->null, 14->null, null],
return [null, 9->null, null, null, null, 21->null, 14->null, null]
解题思路
题目是实现扩建哈希表容量。
哈希表本质上是一个装有LinkedList的数组。新的哈希表是原来哈希表的两倍。遍历原有的数组每个位置，并垂直方向上找到该相同位置上所有的ListNode，并计算该ListNode在新的哈希表数组中的位置。在新的哈希表数组位置，如果已经放入另外的新ListNode了则垂直方向链接ListNode，没有其他的ListNode，则是该位置的第一个新ListNode。
参考代码
/**
* 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 null;
}
int newCapacity = hashTable.length * 2;
ListNode[] newHashTable = new ListNode[newCapacity];
for (int i = 0; i < hashTable.length; i++) {
while (hashTable[i] != null) {
int newPosition = (hashTable[i].val % newCapacity + newCapacity) % newCapacity;
if (newHashTable[newPosition] != null) {
ListNode head = newHashTable[newPosition];
while (head.next != null) {
head = head.next;
}
head.next = new ListNode(hashTable[i].val);
} else {
newHashTable[newPosition] = new ListNode(hashTable[i].val);
}
hashTable[i] = hashTable[i].next;
}
}
return newHashTable;
}
}