706 Design HashMap

706. Design HashMap

1. Question

Design a HashMap without using any built-in hash table libraries.

To be specific, your design should include these functions:

• put(key, value): Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.

• get(key): Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.

• remove(key): Remove the mapping for the value key if this map contains the mapping for the key. Example:

MyHashMap hashMap = new MyHashMap();
hashMap.put(1, 1);          
hashMap.put(2, 2);         
hashMap.get(1);            // returns 1
hashMap.get(3);            // returns -1 (not found)
hashMap.put(2, 1);          // update the existing value
hashMap.get(2);            // returns 1 
hashMap.remove(2);          // remove the mapping for 2
hashMap.get(2);            // returns -1 (not found)

Note:

• All keys and values will be in the range of[0, 1000000].

• The number of operations will be in the range of [1, 10000].

• Please do not use the built-in HashMap library.

2. Implementation

(1) Array + Double Linked List

class MyHashMap {
    final int LEN = 10000;
    ListNode[] buckets;

    /** Initialize your data structure here. */
    public MyHashMap() {
        buckets = new ListNode[LEN];
    }

    /** value will always be non-negative. */
    public void put(int key, int value) {
        int index = getIndex(key);

        if (buckets[index] == null) {
            buckets[index] = new ListNode(-1, -1);
        }

        ListNode preNode = find(buckets[index], key);

        if (preNode.next == null) {
            preNode.next = new ListNode(key, value);
        }
        else {
            preNode.next.val = value;
        }
    }

    /** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
    public int get(int key) {
        int index = getIndex(key);

        if (buckets[index] == null) {
            return -1;
        }

        ListNode preNode = find(buckets[index], key);

        if (preNode.next == null) {
            return -1;
        }
        else {
            return preNode.next.val;
        }
    }

    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    public void remove(int key) {
        int index = getIndex(key);

        if (buckets[index] == null) {
            return;
        }

        ListNode preNode = find(buckets[index], key);

        if (preNode.next == null) {
            return;
        }
        else {
            preNode.next = preNode.next.next;
        }
    }

    class ListNode {
        int key, val;
        ListNode next;

        public ListNode(int key, int value) {
            this.key = key;
            this.val = value;
            next = null;
        }
    }

    public int getIndex(int key) {
        return Integer.hashCode(key) % LEN;
    }

    public ListNode find(ListNode bucket, int key) {
        ListNode curNode = bucket;
        ListNode preNode = null;

        while (curNode != null && curNode.key != key) {
            preNode = curNode;
            curNode = curNode.next;
        }
        return preNode;
    }
}

/**
 * Your MyHashMap object will be instantiated and called as such:
 * MyHashMap obj = new MyHashMap();
 * obj.put(key,value);
 * int param_2 = obj.get(key);
 * obj.remove(key);
 */

3. Time & Space Complexity

时间复杂度: put: O(1), get: O(1), remove: O(1). 空间复杂度: O(n), n为key的个数