# 705.Design Hashset

**705.Design Hashset**

难度:Easy

> 不使用任何内建的哈希表库设计一个哈希集合

具体地说，你的设计应该包含以下的功能

add(value)：向哈希集合中插入一个值。 contains(value) ：返回哈希集合中是否存在这个值。 remove(value)：将给定值从哈希集合中删除。如果哈希集合中没有这个值，什么也不做。

```
示例:

MyHashSet hashSet = new MyHashSet();
hashSet.add(1);         
hashSet.add(2);         
hashSet.contains(1);    // 返回 true
hashSet.contains(3);    // 返回 false (未找到)
hashSet.add(2);          
hashSet.contains(2);    // 返回 true
hashSet.remove(2);          
hashSet.contains(2);    // 返回  false (已经被删除)
```

注意：

所有的值都在 \[1, 1000000]的范围内。 操作的总数目在\[1, 10000]范围内。 不要使用内建的哈希集合库。

直接使用数组索引完成。

```
class MyHashSet {
private:
    bool hset[1000001]={false};
public:
    /** Initialize your data structure here. */
    MyHashSet() {
       
    }
    
    void add(int key) {
        hset[key]=1;
    }
    
    void remove(int key) {
        hset[key]=0;
    }
    
    /** Returns true if this set contains the specified element */
    bool contains(int key) {
        return hset[key];
    }
};

/**
 * 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);
 */
```

> 执行用时 : 212 ms, 在Design HashSet的C++提交中击败了61.47% 的用户\
> 内存消耗 : 66.6 MB, 在Design HashSet的C++提交中击败了31.36% 的用户


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