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.

class MyHashMap {

    int[] map;
    /** Initialize your data structure here. */
    public MyHashMap() {
        map = new int[1000001];
    }
    
    /** value will always be non-negative. */
    public void put(int key, int value) {
        map[key] = value + 1;
    }
    
    /** 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) {
       return map[key] - 1; 
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    public void remove(int key) {
        map[key] = 0;
    }
}

用large size array，很快啊，也很偷懒。

hashmap原理就是用array和linkedlist。

put就是先表示成ListNode，然后通过hash来找数组中key对应的下标，找到后看这个数组有没有linkedlist，如果有就查有没有同样的key有就替换，没有就添加到最后一个。

get是用hash key找数组的下标，如果这个数组下标对应的没有就返回null，有就遍历linkedlist，找到key对应的value。

 

 https://baijiahao.baidu.com/s?id=1665667572592680093&wfr=spider&for=pc

class MyHashMap {
        final ListNode[] nodes = new ListNode[10000];

        public void put(int key, int value) {
            int i = idx(key);
            if (nodes[i] == null)
                nodes[i] = new ListNode(-1, -1);
            ListNode prev = find(nodes[i], key);
            if (prev.next == null)
                prev.next = new ListNode(key, value);
            else prev.next.val = value;
        }

        public int get(int key) {
            int i = idx(key);
            if (nodes[i] == null)
                return -1;
            ListNode node = find(nodes[i], key);
            return node.next == null ? -1 : node.next.val;
        }

        public void remove(int key) {
            int i = idx(key);
            if (nodes[i] == null) return;
            ListNode prev = find(nodes[i], key);
            if (prev.next == null) return;
            prev.next = prev.next.next;
        }

        int idx(int key) { return Integer.hashCode(key) % nodes.length;}

        ListNode find(ListNode bucket, int key) {
            ListNode node = bucket, prev = null;
            while (node != null && node.key != key) {
                prev = node;
                node = node.next;
            }
            return prev;
        }

        class ListNode {
            int key, val;
            ListNode next;

            ListNode(int key, int val) {
                this.key = key;
                this.val = val;
            }
        }
    }