706. Design a Hash Map

Design a hash map (HashMap) without using any built-in hash table library.

Realize MyHashMapcategories:

MyHashMap() Initialize the object with an empty map
void put(int key, int value)To HashMapinsert a key-value pair (key, value). If keyalready exists in the map, the corresponding value is updated value.
int get(int key)Return a specific keymapped value; if the map is not included in keythe map, return -1.
void remove(key)If there is a mapping of keythe mapping is removed keyand its corresponding value.

Example:

输入：
["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
输出：
[null, null, null, 1, -1, null, 1, null, -1]

解释：
MyHashMap myHashMap = new MyHashMap();
myHashMap.put(1, 1); // myHashMap 现在为 [[1,1]]
myHashMap.put(2, 2); // myHashMap 现在为 [[1,1], [2,2]]
myHashMap.get(1);    // 返回 1 ，myHashMap 现在为 [[1,1], [2,2]]
myHashMap.get(3);    // 返回 -1（未找到），myHashMap 现在为 [[1,1], [2,2]]
myHashMap.put(2, 1); // myHashMap 现在为 [[1,1], [2,1]]（更新已有的值）
myHashMap.get(2);    // 返回 1 ，myHashMap 现在为 [[1,1], [2,1]]
myHashMap.remove(2); // 删除键为 2 的数据，myHashMap 现在为 [[1,1]]
myHashMap.get(2);    // 返回 -1（未找到），myHashMap 现在为 [[1,1]]

prompt:

0 <= key, value <= 10^6
Call most 10^4times put, getand removemethods

answer

Simple and crude method:

class MyHashMap {
    
    
private:
    vector<bool> keys;
    vector<int> vals;
public:
    /** Initialize your data structure here. */
    MyHashMap() {
    
    
        keys.resize(1e6 + 1, false);
        vals.resize(1e6 + 1, -1);
    }
    
    /** value will always be non-negative. */
    void put(int key, int value) {
    
    
        keys[key] = true;
        vals[key] = value;
    }
    
    /** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
    int get(int key) {
    
    
        return vals[key];
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    void remove(int key) {
    
    
        keys[key] = false;
        vals[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);
 */

Like 705. Designing a hash set , the chain address method of the official problem solution is adopted:

class MyHashMap {
    
    
private:
    vector<list<pair<int, int>>> map;
    static const int base = 769;
    static int hash(int key) {
    
    
        return key % base;
    }
public:
    /** Initialize your data structure here. */
    MyHashMap() {
    
    
        map.resize(base);
    }
    
    /** value will always be non-negative. */
    void put(int key, int value) {
    
    
        int h = hash(key);
        for(auto it = map[h].begin(); it != map[h].end(); it++){
    
    
            if(it->first == key){
    
    
                it->second = value;
                return;
            }
        }
        map[h].emplace_back(key, value);
    }
    
    /** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
    int get(int key) {
    
    
        int h = hash(key);
        for(auto it = map[h].begin(); it != map[h].end(); it++){
    
    
            if(it->first == key)
                return it->second;
        }
        return -1;
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    void remove(int key) {
    
    
        int h = hash(key);
        for(auto it = map[h].begin(); it != map[h].end(); it++){
    
    
            if(it->first == key){
    
    
                map[h].erase(it);
                return;
            }
        }
    }
};

/**
 * 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);
 */

706. Design a Hash Map

answer

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