Leetcode 705. Design hash collection

Design a hash set without using any built-in hash table Library （HashSet）.

Realization MyHashSet class ：

• void add(key) Insert a value into the hash collection key .
• bool contains(key) Returns whether the value exists in the hash set key .
• void remove(key) Set the given value key Remove... From the hash set . If there is no such value in the hash set , Do nothing .

Example ：

 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]

 explain ：
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 ,（ Removed ）
 

Tips ：

• 0 <= key <= 10^6
• Call at most 10^4 Time add、remove and contains

Code:

class MyHashSet {
    
public:
    MyHashSet() {
    

    }
    
    void add(int key) {
    
        int cnt=count(vec.begin(),vec.end(),key);
        if(!cnt)
        vec.push_back(key);
    }
    
    void remove(int key) {
    
        vector<int>::iterator it=find(vec.begin(),vec.end(),key);
        if(it!=vec.end())
        {
    
            vec[it-vec.begin()]=-1;
        }

    }
    
    bool contains(int key) {
    
     
        return count(vec.begin(),vec.end(),key)!=0;
    }
    private:
    vector<int>vec;
};

/** * 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); */