[leetCode/JS] 706. Design HashMap

난이도 [ 😊 ] Easy

문제 설명

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.

입출력 예

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.

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내 솔루션

• 음.. 그냥 한번 알면 쉬운편, Map이나 Set은 한번 공부해두면 매우 편함.

var MyHashMap = function() {    
    this.map = new Map()
};

/** 
 * @param {number} key 
 * @param {number} value
 * @return {void}
 */
MyHashMap.prototype.put = function(key, value) {
    this.map.set(key, value)
};

/** 
 * @param {number} key
 * @return {number}
 */
MyHashMap.prototype.get = function(key) {
    return this.map.has(key) ? this.map.get(key) : -1
};

/** 
 * @param {number} key
 * @return {void}
 */
MyHashMap.prototype.remove = function(key) {
    this.map.delete(key)
};

감상평