LintCode 129. Rehashing 原创Java参考解答

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

The hash function is:

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

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

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