Description:
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:
// 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();
Link:
https://leetcode.com/problems/insert-delete-getrandom-o1/#/description
解題方法：
哈希表可以O(shè)(1)時(shí)間增加刪除稳摄，vector可以在O(1)時(shí)間取得隨機(jī)數(shù)、刪除尾部元素覆履、在尾部增加元素。
如果用哈希表記錄vector里面元素的位置，在刪除的時(shí)候?qū)ector需要?jiǎng)h除的元素與其尾部互換，就可以在O(1)時(shí)間刪除該元素换吧。
Time Complexity：
O(1)
完整代碼：
class RandomizedSet
{
private:
vector<int> cache;
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;
cache.push_back(val);
hash[val] = cache.size() - 1;
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;
hash[cache.back()] = hash[val]; //交換位置之前哈希表中對(duì)應(yīng)的位置也要交換
swap(cache[cache.size() - 1], cache[hash[val]]);
cache.pop_back();
hash.erase(val);
return true;
}
/** Get a random element from the set. */
int getRandom()
{
return cache[rand()%cache.size()];
}
};
