题目描述
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.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();
题目大意:
设计一个数据结构,在O(1) 的时间完成上面的操作。
思路:
删除时,将最后一个元素移到被删除位置,保证index连续,随机时概率相等。
代码:
class RandomizedSet {
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 (key2index.count(val) != 0)
return false;
key2index[val] = num;
index2key[num] = val;
num += 1;
return true;
}
/** Removes a value from the set. Returns true if the set contained the specified element. */
bool remove(int val) {
if (key2index.count(val) == 0)
return false;
int index1 = key2index[val];
num -= 1;
int index2 = num;
int val2 = index2key[num];
key2index[val2] = index1;
index2key[index1] = val2;
key2index.erase(val);
index2key.erase(index2);
return true;
}
/** Get a random element from the set. */
int getRandom() {
if (num == 0)
return NULL;
int index = (int) (rand() % (num-1)) + 1;
return index2key[index];
}
public:
map<int, int>key2index;
map<int, int>index2key;
int num = 1;
};