Rehashing解题报告

Description:

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]

Link:

http://www.lintcode.com/en/problem/rehashing/

Tips:

在重新计算key的时候，如果value是一个负数，那么使用value % capacity这个公式会产生负数，所以需要使用代替公式：
(value % capacity + capacity) % capacity

完整代码：

vector rehashing(vector hashTable) { int capacity = hashTable.size()*2; vector result(capacity, NULL); for(int i = 0; i < hashTable.size(); i++) { if(hashTable[i] == NULL) continue; else { ListNode* head = hashTable[i]; while(head != NULL) { hashadd(result, head->val); head = head->next; } } } return result; } void hashadd(vector &result, int val) { int key = hashcode(val, result.size()); if(result[key] == NULL) { ListNode* temp = new ListNode(val); result[key] = temp; } else { ListNode* head = result[key]; while(head->next != NULL) head = head->next; ListNode* temp = new ListNode(val); head->next = temp; } } int hashcode(int key, int capacity) { return (key % capacity + capacity) % capacity; }