• LeetCode-381 Insert Delete GetRandom O(1)


    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.

    题目大意

    设计一个数据结构,实现以下三种操作:

      1. 插入:如果插入一个已存在的数字则返回false,否则插入该数字返回true;

      2. 删除:如果删除一个不存在的数字返回false,否则删除该数字返回true;

      3. 获得随机数:随机获得已插入的数字其中的一个,保证所有数字都有几率获得。

    (PS:插入的数字可以重复。)

    示例

    E1

    // 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();

    解题思路

    方法类似LeetCode-381,将map<int,int>换成unordered_map<int,unordered_set<int>>即可以。

    复杂度分析

    时间复杂度:O(1)

    空间复杂度:O(N)

    代码

    class RandomizedCollection {
    private:
        vector<int> nums;
        unordered_map<int, unordered_set<int> > index;
        
    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) {
            nums.emplace_back(val);
            index[val].insert(nums.size() - 1);
            return index[val].size() == 1;
        }
        
        /** Removes a value from the collection. Returns true if the collection contained the specified element. */
        bool remove(int val) {
            auto it = index.find(val);
            if(it != index.end()) {
                auto tmp = *it->second.begin();
                it->second.erase(it->second.begin());
                nums[tmp] = nums.back();
                index[nums.back()].insert(tmp);
                index[nums.back()].erase(nums.size() - 1);
                nums.pop_back();
                if(index[val].empty())
                    index.erase(val);
                return true;
            }
            
            return false;
        }
        
        /** Get a random element from the collection. */
        int getRandom() {
            return nums[rand() % nums.size()];
        }
    };
    
    /**
     * Your RandomizedCollection object will be instantiated and called as such:
     * RandomizedCollection* obj = new RandomizedCollection();
     * bool param_1 = obj->insert(val);
     * bool param_2 = obj->remove(val);
     * int param_3 = obj->getRandom();
     */
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  • 原文地址:https://www.cnblogs.com/heyn1/p/11263940.html
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