[数据结构与算法]Leetcode 380.O(1) 时间插入、删除和获取随机元素

原题链接：Leetcode 380. Insert Delete GetRandom O(1)

Implement the RandomizedSet class:

• RandomizedSet() Initializes the RandomizedSet object.
• bool insert(int val) Inserts an item val into the set if not present. Returns true if the item was not present, false otherwise.
• bool remove(int val) Removes an item val from the set if present. Returns true if the item was present, false otherwise.
• int getRandom() Returns a random element from the current set of elements (it’s guaranteed that at least one element exists when this method is called). Each element must have the same probability of being returned.

You must implement the functions of the class such that each function works in average O(1) time complexity.

Example 1:

Input

["RandomizedSet", "insert", "remove", "insert", "getRandom", "remove", "insert", "getRandom"]
[[], [1], [2], [2], [], [1], [2], []]

Output

[null, true, false, true, 2, true, false, 2]

Explanation
RandomizedSet randomizedSet = new RandomizedSet();
randomizedSet.insert(1); // Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomizedSet.remove(2); // Returns false as 2 does not exist in the set.
randomizedSet.insert(2); // Inserts 2 to the set, returns true. Set now contains [1,2].
randomizedSet.getRandom(); // getRandom() should return either 1 or 2 randomly.
randomizedSet.remove(1); // Removes 1 from the set, returns true. Set now contains [2].
randomizedSet.insert(2); // 2 was already in the set, so return false.
randomizedSet.getRandom(); // Since 2 is the only number in the set, getRandom() will always return 2.

Constraints:

• -2³¹ <= val <= 2³¹ - 1
• At most 2 * 10⁵ calls will be made to insert, remove, and getRandom.
• There will be at least one element in the data structure when getRandom is called.

方法一：vector+哈希表

思路：

本题目的要求是实现一个类 RandomizedSet，并且它的插入、删除、获取随机元素的时间复杂度都是O(1)。

• unordered_map<int,int> mp 建立数组 nums[]元素的值与序号idx的映射
• insert(x) : 如果 x 不在 mp 中，就插入数组nums和哈希表mp中
• remove(x) : 如果 x 在 mp 中 :
  先找到需要删除的元素的对应数组中的索引 index = mp[x]
  然后与数组最后一个元素交换 , 然后删除数组的最后一个元素即可 (top 表示当前数组中有的元素个数)
• getRandom() : 直接随机一个 [0,top) 中的下标即可

c++代码：

class RandomizedSet {
private:
    // nums存放值 mp存放<值, 序号>
    vector<int> nums;
    unordered_map<int, int> mp;

public:
    RandomizedSet() {
        // 初始化随机数种子
        srand((unsigned)time(NULL));
    }
    
    // 插入
    bool insert(int val) {
        if (indices.count(val)) {
            return false;
        }
        int index = nums.size();
        nums.emplace_back(val);
        mp[val] = index;
        return true;
    }
    
    // 删除
    bool remove(int val) {
        if (!mp.count(val)) {
            return false;
        }
        // 把要删除的和数组的最后一个进行交换 相当于删除最后一个
        int index = mp[val];
        int last = nums.back();
        nums[index] = last;
        mp[last] = index;
        nums.pop_back();
        mp.erase(val);
        return true;
    }
    
    // 随机返回一项
    int getRandom() {
        int randomIndex = rand()%nums.size();
        return nums[randomIndex];
    }

};

/**
 * Your RandomizedSet object will be instantiated and called as such:
 * RandomizedSet* obj = new RandomizedSet();
 * bool param_1 = obj->insert(val);
 * bool param_2 = obj->remove(val);
 * int param_3 = obj->getRandom();
 */

复杂度分析：

• 时间复杂度：
  • 插入：O(1)
  • 删除：O(1)
  • 获取：O(1)
• 空间复杂度：O(n)，为数组和哈希表的开销

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