# LeetCode Insert Delete GetRandom O(1)

LeetCode Insert Delete GetRandom O(1)

Design a data structure that supports all following operations in average O(1) time.

1. `insert(val)`: Inserts an item val to the set if not already present.
2. `remove(val)`: Removes an item val from the set if present.
3. `getRandom`: Returns a random element from current set of elements. Each element must have the same probability of being returned.

Example:

```// Init an empty set.
RandomizedSet randomSet = new RandomizedSet();

// Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomSet.insert(1);

// Returns false as 2 does not exist in the set.
randomSet.remove(2);

// Inserts 2 to the set, returns true. Set now contains [1,2].
randomSet.insert(2);

// getRandom should return either 1 or 2 randomly.
randomSet.getRandom();

// Removes 1 from the set, returns true. Set now contains [2].
randomSet.remove(1);

// 2 was already in the set, so return false.
randomSet.insert(2);

// Since 2 is the only number in the set, getRandom always return 2.
randomSet.getRandom();```

```class RandomizedSet {
private:
vector<int> nums;
unordered_map<int, int> hash;
public:
/** Initialize your data structure here. */
RandomizedSet() {
}

/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
bool insert(int val) {
if (hash.find(val) != hash.end())return false;
hash[val] = nums.size();
nums.push_back(val);
return true;
}

/** Removes a value from the set. Returns true if the set contained the specified element. */
bool remove(int val) {
if (hash.find(val) == hash.end())return false;
int pos = hash[val];
swap(nums[pos], nums[nums.size() - 1]);
//下面两句顺序不能乱，因为有可能删除的就是最后一个元素
hash[nums[pos]] = pos;
hash.erase(val);
nums.pop_back();
return true;
}

/** Get a random element from the set. */
int getRandom() {
return nums[rand() % nums.size()];
}
};
```