LeetCode Insert Delete GetRandom O(1) 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.
// 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();
本题要求实现这样一种数据结构,插入、删除和产生集合中的一个随机数的时间复杂度都是O(1)。 插入和删除要是O(1),可以借助Hash,但Hash表不能以O(1)时间生成随机数。如果把数存在数组中,则能以O(1)的时间生成随机数,但数组的删除不能O(1)。所以可以把Hash和数组结合起来。 数字存储在数组中,Hash表中存储数字在数组中的下标。插入时,插入到数组末尾,同时更新Hash表中的下标。删除时,把数字和数组末尾的数字交换,这样删除数组末尾元素可以用O(1)时间完成,同时也要把Hash表中的下标抹掉,并更新原数组最后一个元素的下标。产生随机数就好办了,知道数组长度,rand下标就好。 完整代码如下: [cpp] class RandomizedSet { private: vector<int> nums; unordered_map<int, int> hash; public: /** Initialize your data structure here. */ RandomizedSet() { } /** Inserts a value to the set. Returns true if the set did not already contain the specified element. */ bool insert(int val) { if (hash.find(val) != hash.end())return false; hash[val] = nums.size(); nums.push_back(val); return true; } /** Removes a value from the set. Returns true if the set contained the specified element. */ bool remove(int val) { if (hash.find(val) == hash.end())return false; int pos = hash[val]; swap(nums[pos], nums[nums.size() – 1]); //下面两句顺序不能乱,因为有可能删除的就是最后一个元素 hash[nums[pos]] = pos; hash.erase(val); nums.pop_back(); return true; } /** Get a random element from the set. */ int getRandom() { return nums[rand() % nums.size()]; } }; [/cpp] 本代码提交AC,用时59MS。]]>
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