381. Insert Delete GetRandom O(1) - Duplicates allowed

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

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();
这里的结构允许出现重复的数字,且在随机选取某个数时,概率相等.
仍然用hash表，但记录val在数组中的位置用set。
class RandomizedCollection {   map<int,set<int>> locs;//val to loc    vector<int> nums;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.push_back(val);        locs[val].insert(nums.size()-1);        if(locs[val].size()==1)        return 1;        else return 0;    }        /** Removes a value from the collection. Returns true if the collection contained the specified element. */    bool remove(int val) {       if(locs.find(val)==locs.end()) return 0;       int idx=*locs[val].begin();       int last=nums[nums.size()-1];       nums[idx]=last;              locs[val].erase(idx);//这里必须先删除一个才可以插入，如果last和val相等，先插入，会没有效果。这里调试了好久才找到错误       locs[last].insert(idx);       locs[last].erase(nums.size()-1);       nums.pop_back();                   if(locs[val].size()==0) locs.erase(val);       return 1;          }        /** Get a random element from the collection. */    int getRandom() {      return nums[rand()%nums.size()];      }};