[LeetCode August Challange]Day2-Design HashSet

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

To be specific, your design should include these functions:

• add(value): Insert a value into the HashSet.
• contains(value): Return whether the value exists in the HashSet or not.
• remove(value): Remove a value in the HashSet. If the value does not exist in the HashSet, do nothing.

Example:

MyHashSet hashSet = new MyHashSet();
hashSet.add(1);         
hashSet.add(2);         
hashSet.contains(1);    // returns true
hashSet.contains(3);    // returns false (not found)
hashSet.add(2);          
hashSet.contains(2);    // returns true
hashSet.remove(2);          
hashSet.contains(2);    // returns false (already removed)

Note:

• All values will be in the range of [0, 1000000].
• The number of operations will be in the range of [1, 10000].
• Please do not use the built-in HashSet library.

---

solution

class MyHashSet {
public:
    /** Initialize your data structure here. */
    MyHashSet() {
        content = vector<bool>(1000001, false);
    }
    
    void add(int key) {
        content[key] = true;
    }
    
    void remove(int key) {
        content[key] = false;
    }
    
    /** Returns true if this set contains the specified element */
    bool contains(int key) {
        return content[key];
    }
private:
    vector<bool> content;
};

/**
 * Your MyHashSet object will be instantiated and called as such:
 * MyHashSet* obj = new MyHashSet();
 * obj->add(key);
 * obj->remove(key);
 * bool param_3 = obj->contains(key);
 */

Hash table的概念圖：

理論上輸入(key, value)，key經過hash function之後，會去存取相對應index的內容。
table得看需求設計大小，或是進階一點，動態大小。
(index, content) pair 後面的content也可延伸為linked list。

按照此題需求，index範圍是[0, 1000000]，也只需知道此數值是否有存起來，因此hash function f(x) = x，content 就存有沒有存入此數值即可。