381. Insert Delete GetRandom O - Duplicates allowed

Posted 2021-03-01 habibah-chang

问题：

设计数据结构，使得以下三个方法的时间复杂度都为O(1) 

允许插入重复数字。


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();

　　

解法：

本题需让insert，remove的复杂度为O(1)

数组的insert方法复杂度满足，remove中的查找不满足，因此加入hash来帮助记录消化查找的代价。

另，由于允许可重复的数字，那么再使hash中的value为一个数组，记录同一个数字的不同位置

vector<int, int> nums //fisrt: 插入数值，second：该数字在map中，同一个数字数组中存的位置 i1

unordered_map<int, vector<int>>  //(key: nums[i], value: [i1,i2,i3...])

代码参考：

 1 class RandomizedCollection {
 2 public:
 3     /** Initialize your data structure here. */
 4     RandomizedCollection() {
 5         
 6     }
 7     
 8     /** Inserts a value to the collection. Returns true if the collection did not already contain the specified element. */
 9     bool insert(int val) {
10         
11         if(mapnums.find(val)==mapnums.end()){
12             mapnums[val].push_back(nums.size());
13             nums.push_back({val,0});
14         }else{
15             vector<int>* tmp = &mapnums[val];
16             tmp->push_back(nums.size());
17             nums.push_back({val,tmp->size()-1});
18         }
19         return true;
20     }
21     
22     /** Removes a value from the collection. Returns true if the collection contained the specified element. */
23     bool remove(int val) {
24         if(mapnums.find(val)==mapnums.end()) return false;
25         pair<int,int> last = nums.back();
26         int indexval = mapnums[val].back();
27         nums[indexval]=last;
28         mapnums[last.first][last.second]=indexval;
29         mapnums[val].pop_back();
30         if(mapnums[val].empty()){
31             mapnums.erase(val);
32         }
33         nums.pop_back();
34         return true;
35     }
36     
37     /** Get a random element from the collection. */
38     int getRandom() {
39         return nums[rand() % nums.size()].first;
40     }
41 private:
42     vector<pair<int, int>> nums;
43     unordered_map<int,vector<int>> mapnums;
44 };
45 
46 /**
47  * Your RandomizedCollection object will be instantiated and called as such:
48  * RandomizedCollection* obj = new RandomizedCollection();
49  * bool param_1 = obj->insert(val);
50  * bool param_2 = obj->remove(val);
51  * int param_3 = obj->getRandom();
52  */