LintCode 129. Rehashing (链表和指针结合好题)

129. Rehashing
     中文English
     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
Example 1

Input : [null, 21->9->null, 14->null, null]
Output : [null, 9->null, null, null, null, 21->null, 14->null, null]
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.

解法1：注意有两个循环，在新数组和旧数组中都要循环
代码如下：

/**
 * 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 rehashing(vector hashTable) {
        int capacity = hashTable.size(); 
        int capacity2 = capacity << 1;
        vector result(capacity2, NULL);
        for (int i = 0; i < capacity; ++i) {
            ListNode * curNode = hashTable[i];
            while(curNode) {
                int val = curNode->val;
                ListNode * newNode = new ListNode(val);
                int newIndex = val < 0 ? (val % capacity2 + capacity2) % capacity2 : val % capacity2;
                if (result[newIndex] == NULL) {
                    result[newIndex] = newNode;
                } else {
                    ListNode * tmp = result[newIndex];
                    while(tmp->next) {
                        tmp = tmp->next;
                    }
                    tmp->next = newNode;
                }
                curNode = curNode->next;
            }
        }
        
        return result;
    }
};