leetcode 380

问题

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:

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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();

分析

使用一个vector来保存每个值,使用hashmap保存val到index的映射关系,同时可以在线性时间内判断出一个元素是否存在。

另:在没有要求元素顺序位置的情况下,使用线性表也可以在O(1)时间内插入元素和在O(1)时间内删除执行元素(但要在知道其下标的前提下,将该元素与vector末尾的元素进行交换,删除末尾元素)。

代码

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class RandomizedSet {
public:
/** Initialize your data structure here. */
unordered_map<int,int> mp; // val -> index
vector<int> a;
random_device rd;
RandomizedSet() {
}
/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
bool insert(int val) {
if (mp.find(val) != mp.end())
return false;
a.push_back(val);
mp[val] = a.size()-1;
return true;
}
/** Removes a value from the set. Returns true if the set contained the specified element. */
bool remove(int val) {
if (mp.find(val) == mp.end())
return false;
int index = mp[val];
if (index < a.size()-1) {
mp[a.back()] = index;
a[index] = a.back();
}
mp.erase(val);
a.pop_back();
return true;
}
/** Get a random element from the set. */
int getRandom() {
return a[rd()%a.size()];
}
};