#129 Rehashing

最新推荐文章于 2024-10-23 18:18:52 发布

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#lintcode #hash #linked list

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本文介绍了一种哈希表在达到特定负载因子时进行扩容的方法，通过双倍增加容量并重新散列所有键来解决冲突问题。文章提供了一个具体的示例，展示了如何将一个包含碰撞的哈希表通过重新散列变成一个更大且分布更均匀的哈希表。

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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 .

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.

Have you met this question in a real interview?

Yes

Example

Given [null, 21->9->null, 14->null, null],

return[null, 9->null, null, null, null, 21->null, 14->null, null]

题目思路：

这题需要注意的是每个位置上可能是个linked list，所以从旧位置取出node时，还是需要保持剩下的linked list在旧位置上。把node放进新位置时，需要把它放在新位置的linked list的尾巴上。

Mycode（AC = 23ms）：

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @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
     */    
    vector<ListNode*> rehashing(vector<ListNode*> hashTable) {
        // write your code here
        vector<ListNode*> rehashTable(hashTable.size() * 2, NULL);
        int new_size = hashTable.size() * 2;
        
        for (int i = 0; i < hashTable.size(); i++) {
            if (hashTable[i] != NULL) {
                // take out the head node of hashTable[i]
                ListNode *head = hashTable[i], *next = head->next;
                hashTable[i] = next;
                head->next = NULL;
                
                // compute the new pos, and put the node to the new pos
                // or end of linked list at new pos
                int pos = (head->val % new_size + new_size) % new_size;
                if (rehashTable[pos] == NULL) {
                    rehashTable[pos] = head;
                }
                else {
                    ListNode *node = rehashTable[pos];
                    while (node->next) {
                        node = node->next;
                    }
                    node->next = head;
                }
                
                i--;
            }
        }
        
        return rehashTable;
    }
};