Design a data structure that supports all following operations in average O(1) time.
insert(val)
: Inserts an item val to the set if not already present.
remove(val)
: Removes an item val from the set if present.
getRandom
: Returns a random element from current set of elements. Each element must have the same probability of being returned.
Example:
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| // Init an empty set. RandomizedSet randomSet = new RandomizedSet(); // Inserts 1 to the set. Returns true as 1 was inserted successfully. randomSet.insert(1); // Returns false as 2 does not exist in the set. randomSet.remove(2); // Inserts 2 to the set, returns true. Set now contains [1,2]. randomSet.insert(2); // getRandom should return either 1 or 2 randomly. randomSet.getRandom(); // Removes 1 from the set, returns true. Set now contains [2]. randomSet.remove(1); // 2 was already in the set, so return false. randomSet.insert(2); // Since 2 is the only number in the set, getRandom always return 2. randomSet.getRandom();
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思路
利用hashtable,访问与删除平均都可以达到O(1);
利用vector记录,random访问平均是O(1),在头尾插入删除平均访问时间也是O(1);
vector在删除的时候,可以只删除最后一个元素,然后把该元素和待删除的元素位置对换即可。这样的时间复杂度也是O(1)。
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| class RandomizedSet { private: unordered_map<int, int> map; vector<int> values; public: RandomizedSet() { } bool insert(int val) { if(map.find(val) == map.end()){ map[val] = values.size(); values.push_back(val); return true; } else{ return false; } } bool remove(int val) { if(map.find(val) == map.end()){ return false; } else{ int index = map[val]; int lastele = values.back(); if(index == values.size() - 1){ map.erase(val); values.pop_back(); } else{ map.erase(val); map[lastele] = index; values[index] = lastele; values.pop_back(); } return true; } } int getRandom() { if(values.size() > 0){ int index = rand() % values.size(); int choose = values[index]; return choose; } else{ return -1; } } }; * 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(); */
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