• Lintcode: 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->9->null, 14->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 table: [null, 9, null, null, null, 21, 14, null] Given the original hash table, return the new hash table after rehashing . Note For negative integer in hash table, the position can be calculated as follow: In 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. In 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]

    这道题就是根据条件老老实实的做

    /**
     * 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) {
            // write your code here
            int oldSize = hashTable.length;
            int newSize = oldSize * 2;
            if (hashTable==null || oldSize==0) return null;
            ListNode[] res = new ListNode[newSize];
            for (int i=0; i<oldSize; i++) {
                if (hashTable[i] != null) rehash(hashTable, res, i);
            }
            return res;
        }
        
        public void rehash(ListNode[] hashTable, ListNode[] res, int i) {
            int newSize = res.length;
            ListNode cur = hashTable[i];
            while (cur != null) {
                int val = cur.val;
                int newPos = val>=0? val%newSize : (val%newSize+newSize)%newSize;
                if (res[newPos] == null) res[newPos] = new ListNode(val);
                else {
                    ListNode temp = res[newPos];
                    while (temp.next != null) {
                        temp = temp.next;
                    }
                    temp.next = new ListNode(val);
                }
                cur = cur.next;
            }
        }
    };
    

      

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  • 原文地址:https://www.cnblogs.com/apanda009/p/7797167.html
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