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

[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; 
    } 
}

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