题目描述

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

Note: Duplicate elements are allowed.
insert(val): Inserts an item val to the collection.
remove(val): Removes an item val from the collection if present.
getRandom: Returns a random element from current collection of elements. The probability of each element being returned is linearly related to the number of same value the collection contains.

样例

Example:

// Init an empty collection.
RandomizedCollection collection = new RandomizedCollection();

// Inserts 1 to the collection. Returns true as the collection did not contain 1.
collection.insert(1);

// Inserts another 1 to the collection. Returns false as the collection contained 1. Collection now contains [1,1].
collection.insert(1);

// Inserts 2 to the collection, returns true. Collection now contains [1,1,2].
collection.insert(2);

// getRandom should return 1 with the probability 2/3, and returns 2 with the probability 1/3.
collection.getRandom();

// Removes 1 from the collection, returns true. Collection now contains [1,2].
collection.remove(1);

// getRandom should return 1 and 2 both equally likely.
collection.getRandom();

算法1

(Hashmap+swap) $O(n^2)$

因为需要O（1）时间复杂度，考虑Lookup table。 由于有重复数据，hashmap存储该数据值以及不同index。删除时swap链表尾部的元素和待删除元素，update其index信息。

时间复杂度

O（1）
O（n）

参考文献

C++ 代码

class RandomizedCollection {
public:
    /** Initialize your data structure here. */
    RandomizedCollection() {

    }

    /** Inserts a value to the collection. Returns true if the collection did not already contain the specified element. */
    bool insert(int val) {
        m[val].push(nums.size());
        nums.push_back(val);
        return m[val].size()==1;
    }

    /** Removes a value from the collection. Returns true if the collection contained the specified element. */
    bool remove(int val) {
        if(m[val].empty()) return false;
        int idx=m[val].top();
        m[val].pop();
        if(nums.size()-1!=idx)
        {
            int t=nums.back();
            nums[idx]=t;
            m[t].pop();
            m[t].push(idx);
        }
        nums.pop_back();
        return true;
    }

    /** Get a random element from the collection. */
    int getRandom() {
        return nums[rand()%nums.size()];
    }
private:
    vector<int> nums;
    unordered_map<int,priority_queue<int>> m;// list begin(), erase
};