#129 Rehashing

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   7hash : [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;    }};