Description
https://leetcode.com/problems/design-hashset/
Design a HashSet without using any built-in hash table libraries.
Implement MyHashSet class:
void add(key)Inserts the valuekeyinto the HashSet.bool contains(key)Returns whether the valuekeyexists in the HashSet or not.void remove(key)Removes the valuekeyin the HashSet. Ifkeydoes not exist in the HashSet, do nothing.
Example 1:
Input ["MyHashSet", "add", "add", "contains", "contains", "add", "contains", "remove", "contains"] [[], [1], [2], [1], [3], [2], [2], [2], [2]] Output
[null, null, null, true, false, null, true, null, false]
Explanation MyHashSet myHashSet = new MyHashSet(); myHashSet.add(1); // set = [1] myHashSet.add(2); // set = [1, 2] myHashSet.contains(1); // return True myHashSet.contains(3); // return False, (not found) myHashSet.add(2); // set = [1, 2] myHashSet.contains(2); // return True myHashSet.remove(2); // set = [1] myHashSet.contains(2); // return False, (already removed)
Constraints:
0 <= key <= 106- At most
104calls will be made toadd,remove, andcontains.
Follow up: Could you solve the problem without using the built-in HashSet library?
Explanation
In Python, we can implement with a list.
Python Solution
class MyHashSet:
def __init__(self):
"""
Initialize your data structure here.
"""
self.hash_set = []
def add(self, key: int) -> None:
if key not in self.hash_set:
self.hash_set.append(key)
def remove(self, key: int) -> None:
if key in self.hash_set:
self.hash_set.remove(key)
def contains(self, key: int) -> bool:
"""
Returns true if this set contains the specified element
"""
return key in self.hash_set
# Your MyHashSet object will be instantiated and called as such:
# obj = MyHashSet()
# obj.add(key)
# obj.remove(key)
# param_3 = obj.contains(key)
- Time Complexity: O(N).
- Space Complexity: O(N).