[LeetCode] 706. Design 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.

设计哈希映射。

题意很简单，不过这道题实现起来，代码量稍微有点多的。我这里给出两种方案。

首先是直接用数组。因为条件里面说了key和value的范围是正数，所以不存在负数的情况，所以可以用数组来处理。再加上数据的范围不是那么大，所以这道题用数组是可以解决的。

class MyHashMap {
    int[] table;

    /** Initialize your data structure here. */
    public MyHashMap() {
        table = new int[1000001];
        Arrays.fill(table, -1);
    }

    /** value will always be non-negative. */
    public void put(int key, int value) {
        table[key] = 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) {
        return table[key];
    }

    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    public void remove(int key) {
        table[key] = -1;
    }
}

/**
 * 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);
 */

再来我提供一个list的做法。需要创建两个list of list，一个用来记录key，一个用来记录value，这样key和value的键值对才能对应起来。这两个list of list的长度我给了500，不是很大，但是解决这个题是够了。当看到一个key的时候，为了避免哈希碰撞，我把这个key % 500。因为我创建的是list of list，所以即使再有冲突也不怕，加到那个对应的sublist里面即可。value加入的位置始终跟key被加入的位置一样。

class MyHashMap {
    List<List<Integer>> keys;
    List<List<Integer>> values;
    int len = 500;

    /** Initialize your data structure here. */
    public MyHashMap() {
        keys = new ArrayList<>();
        values = new ArrayList<>();
        for (int i = 0; i < len; i++) {
            keys.add(i, new ArrayList<>());
            values.add(i, new ArrayList<>());
        }
    }

    private boolean contains(int key) {
        return keys.get(key % len).contains(key);
    }
    
    /** value will always be non-negative. */
    public void put(int key, int value) {
        if (contains(key)) {
            remove(key);
        }
        keys.get(key % len).add(key);
        values.get(key % len).add(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) {
        if (contains(key)) {
            int index = keys.get(key % len).indexOf((Integer) key);
            return values.get(key % len).get(index);
        }
        return -1;
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    public void remove(int key) {
        if (contains(key)) {
            int index = keys.get(key % len).indexOf((Integer) key);
            keys.get(key % len).remove(index);
            values.get(key % len).remove(index);
        }
    }
}

/**
 * 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);
 */

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