Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

[LeetCode] 380. Insert Delete GetRandom O(1) #380

Open
grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 380. Insert Delete GetRandom O(1) #380

grandyang opened this issue May 30, 2019 · 0 comments

Comments

@grandyang
Copy link
Owner

grandyang commented May 30, 2019

 

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 1 is the only number in the set, getRandom always return 1.
randomSet.getRandom();

 

这道题让我们在常数时间范围内实现插入删除和获得随机数操作,如果这道题没有常数时间的限制,那么将会是一道非常简单的题,直接用一个 HashSet 就可以搞定所有的操作。但是由于时间的限制,无法在常数时间内实现获取随机数,所以只能另辟蹊径。此题的正确解法是利用到了一个一维数组和一个 HashMap,其中数组用来保存数字,HashMap 用来建立每个数字和其在数组中的位置之间的映射,对于插入操作,先看这个数字是否已经在 HashMap 中存在,如果存在的话直接返回 false,不存在的话,将其插入到数组的末尾,然后建立数字和其位置的映射。删除操作是比较 tricky 的,还是要先判断其是否在 HashMap 里,如果没有,直接返回 false。由于 HashMap 的删除是常数时间的,而数组并不是,为了使数组删除也能常数级,实际上将要删除的数字和数组的最后一个数字调换个位置,然后修改对应的 HashMap 中的值,这样只需要删除数组的最后一个元素即可,保证了常数时间内的删除。而返回随机数对于数组来说就很简单了,只要随机生成一个位置,返回该位置上的数字即可,参见代码如下:

 

class RandomizedSet {
public:
    RandomizedSet() {}
    bool insert(int val) {
        if (m.count(val)) return false;
        nums.push_back(val);
        m[val] = nums.size() - 1;
        return true;
    }
    bool remove(int val) {
        if (!m.count(val)) return false;
        int last = nums.back();
        m[last] = m[val];
        nums[m[val]] = last;
        nums.pop_back();
        m.erase(val);
        return true;
    }
    int getRandom() {
        return nums[rand() % nums.size()];
    }
private:
    vector<int> nums;
    unordered_map<int, int> m;
};

 

Github 同步地址:

#380

 

类似题目:

Insert Delete GetRandom O(1) - Duplicates allowed

 

参考资料:

https://leetcode.com/problems/insert-delete-getrandom-o1/

https://leetcode.com/problems/insert-delete-getrandom-o1/discuss/85422/AC-C%2B%2B-Solution.-Unordered_map-%2B-Vector

https://leetcode.com/problems/insert-delete-getrandom-o1/discuss/85401/Java-solution-using-a-HashMap-and-an-ArrayList-along-with-a-follow-up.-(131-ms)

 

LeetCode All in One 题目讲解汇总(持续更新中...)

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
Labels
None yet
Projects
None yet
Development

No branches or pull requests

1 participant