705. Design HashSet
Description
Design a HashSet without using any built-in hash table libraries.
Implement MyHashSet
class:
void add(key)
Inserts the valuekey
into the HashSet.bool contains(key)
Returns whether the valuekey
exists in the HashSet or not.void remove(key)
Removes the valuekey
in the HashSet. Ifkey
does 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
104
calls will be made toadd
,remove
, andcontains
.
Solutions
Solution 1: Static Array Implementation
Directly create an array of size $1000001$, initially with each element set to false
, indicating that the element does not exist in the hash set.
When adding an element to the hash set, set the corresponding position in the array to true
; when deleting an element, set the corresponding position in the array to false
; when checking if an element exists, directly return the value at the corresponding position in the array.
The time complexity of the above operations is $O(1)$.
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Solution 2: Array of Linked Lists
We can also create an array of size $SIZE=1000$, where each position in the array is a linked list.
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