[leetcode]706. Design HashMap设计HashMap

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.

 题意：

设计一个HashMap

Solution: LinkedList 

初始化ListNode[]nodes = new ListNode[10000]。 size = 10000是题意要求。 那么，该ListNode[] 每个元素的default值为null。 这也是HashMap相比HashTable的优势：HashMap允许key为null。 

举例，put(2,4),  用hashCode()算出对应在ListNode[]nodes中的坐标 i = 2 

 -->-->

若再put(2,9) , 则因为有相同的key,  之前的(2,4)会被覆盖成(2,9)

若再put(1002, 4)， 用hashCode()算出对应在ListNode[]nodes中的坐标 i = 2。

-->-->

code

class MyHashMap {
        /** Initialize your data structure here. */
        int size = 1000;
        ListNode[] nodes = new ListNode[size];
        /** value will always be non-negative. */
        public void put(int key, int value) {   
            int i = hashId(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;
        }
         /** 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 i = hashId(key);
            if (nodes[i] == null)
                return -1;
            ListNode node = find(nodes[i], key);
            return node.next == null ? -1 : node.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 i = hashId(key);
            if (nodes[i] == null) return;
            ListNode prev = find(nodes[i], key);
            if (prev.next == null) return;
            prev.next = prev.next.next;
        }

        public int hashId(int key) { 
            return Integer.hashCode(key) % size;
        }

        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;
            }
        }
}