Design HashMap[E,Array,Hash Table,Linked List,Design,Hash Function]

https://leetcode.com/problems/design-hashmap/description/

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

Implement the MyHashMap class:

• MyHashMap() initializes the object with an empty map.
• void put(int key, int value) inserts a (key, value) pair into the HashMap. If the key already exists in the map, update the corresponding value.
• int get(int key) returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
• void remove(key) removes the key and its corresponding value if the map contains the mapping for the key.

 

Example 1:

Input
["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
Output
[null, null, null, 1, -1, null, 1, null, -1]

Explanation
MyHashMap myHashMap = new MyHashMap();
myHashMap.put(1, 1); // The map is now [[1,1]]
myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
myHashMap.get(1);    // return 1, The map is now [[1,1], [2,2]]
myHashMap.get(3);    // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
myHashMap.get(2);    // return 1, The map is now [[1,1], [2,1]]
myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
myHashMap.get(2);    // return -1 (i.e., not found), The map is now [[1,1]]

 

Constraints:

• 0 <= key, value <= 106
• At most 104 calls will be made to put, get, and remove.

 

1. Python

 

풀면서도 이게 맞나 싶은 풀이법..

class MyHashMap:
    def __init__(self):
        self.k = []
        self.v = []
    def put(self, key: int, value: int) -> None:
        if key not in self.k:
            self.k.append(key)
            self.v.append(value)
        else:
            idx = self.k.index(key)
            self.v[idx] = value
    def get(self, key: int) -> int:
        if key not in self.k : return -1
        else:
            idx = self.k.index(key)
            return self.v[idx]

    def remove(self, key: int) -> None:
        if key in self.k:
            idx = self.k.index(key)
            self.k.pop(idx)
            self.v.pop(idx)


# Your MyHashMap object will be instantiated and called as such:
# obj = MyHashMap()
# obj.put(key,value)
# param_2 = obj.get(key)
# obj.remove(key)

 

아래는 정석 풀이법

class MyHashMap:

    def __init__(self):
        self.size = 1000
        self.map = [[] for _ in range(self.size)]

    def put(self, key: int, value: int) -> None:
        hash_key = key % self.size
        for pair in self.map[hash_key]:
            if pair[0] == key:
                pair[1] = value
                return
        self.map[hash_key].append([key, value])

    def get(self, key: int) -> int:
        hash_key = key % self.size
        for pair in self.map[hash_key]:
            if pair[0] == key:
                return pair[1]
        return -1

    def remove(self, key: int) -> None:
        hash_key = key % self.size
        for pair in self.map[hash_key]:
            if pair[0] == key:
                self.map[hash_key].remove(pair)
                return

# Your MyHashMap object will be instantiated and called as such:
# obj = MyHashMap()
# obj.put(key, value)
# param_2 = obj.get(key)
# obj.remove(key)

 

 

2. C

 

typedef struct {
    
    int *key;
} MyHashMap;


MyHashMap* myHashMapCreate() {
    
    MyHashMap* ret = (MyHashMap*)malloc(sizeof(MyHashMap));
    ret->key = (int*)malloc(sizeof(int) * 1000001);
    memset(ret->key, -1, sizeof(int) * 1000001);

    return ret;
}

void myHashMapPut(MyHashMap* obj, int key, int value) {
    
    obj->key[key] = value;
}

int myHashMapGet(MyHashMap* obj, int key) {
    
    return obj->key[key];
}

void myHashMapRemove(MyHashMap* obj, int key) {
    
    obj->key[key] = -1;
}

void myHashMapFree(MyHashMap* obj) {
    
    free(obj->key);
    obj->key = NULL;
    free(obj);
    obj = NULL;
}

 

아.. 진짜 짜아승난다!

 

여기까지..