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Kotlin-based LeetCode algorithm problem solutions, regularly updated
706\. Design HashMap
Easy
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`.