LeetCode #381 - Insert Delete GetRandom O(1) - Duplicates allowed

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本文介绍了一种数据结构的设计，该结构支持平均O(1)时间复杂度的插入、删除和随机选择操作，允许元素重复。通过使用哈希表和动态数组，确保了高效的操作执行。

题目描述：

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

Note: Duplicate elements are allowed.

1. insert(val): Inserts an item val to the collection.

2. remove(val): Removes an item val from the collection if present.

3. 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.

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();

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) {
        v.push_back(val);
        if(hash.count(val)==0) 
        {
            hash[val]={v.size()-1};
            return true;
        }
        else 
        {
            hash[val].insert(v.size()-1);
            return false;
        }
    }
    
    /** Removes a value from the collection. Returns true if the collection contained the specified element. */
    bool remove(int val) {
        if(hash.count(val)==0) return false;
        int i=*hash[val].begin();
        //先将hash[val]中的i删除
        if(hash[val].size()==1) hash.erase(hash.find(val));
        else hash[val].erase(hash[val].begin());
        if(i!=v.size()-1)
        {   //如果i等于v.size()-1，那么就不需要再做交换
            int x=v.back();
            //再插入i，将hash[x]中的v.size()-1删除
            hash[x].insert(i);
            hash[x].erase(hash[x].find(v.size()-1));
            swap(v[i],v.back());
        }
        v.pop_back();
        return true;
    }
    
    /** Get a random element from the collection. */
    int getRandom() {
        return v[rand()%v.size()];
    }
    
private:
    unordered_map<int,unordered_set<int>> hash; // 哈希表记录数值和对应的下标集合
    vector<int> v;
};