LeetCode Insert Delete GetRandom O(1)

LeetCode Insert Delete GetRandom O(1)

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

  1. insert(val): Inserts an item val to the set if not already present.
  2. remove(val): Removes an item val from the set if present.
  3. getRandom: Returns a random element from current set of elements. Each element must have the same probability of being returned.

Example:

// 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:

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()];
	}
};

本代码提交AC,用时59MS。

One thought on “LeetCode Insert Delete GetRandom O(1)

  1. Pingback: LeetCode Insert Delete GetRandom O(1) – Duplicates allowed | bitJoy > code

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