题目描述
题目链接:380. Insert Delete GetRandom O(1)
Implement the
RandomizedSet
class:
RandomizedSet()
Initializes theRandomizedSet
object.bool insert(int val)
Inserts an itemval
into the set if not present. Returnstrue
if the item was not present,false
otherwise.bool remove(int val)
Removes an itemval
from the set if present. Returnstrue
if the item was present,false
otherwise.int getRandom()
Returns a random element from the current set of elements (it’s guaranteed that at least one element exists when this method is called). Each element must have the same probability of being returned.You must implement the functions of the class such that each function works in average
O(1)
time complexity.
例子
例子 1
Input:
["RandomizedSet", "insert", "remove", "insert", "getRandom", "remove", "insert", "getRandom"]
[[], [1], [2], [2], [], [1], [2], []]
Output:[null, true, false, true, 2, true, false, 2]
Explanation:RandomizedSet randomizedSet = new RandomizedSet();
randomizedSet.insert(1); // Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomizedSet.remove(2); // Returns false as 2 does not exist in the set.
randomizedSet.insert(2); // Inserts 2 to the set, returns true. Set now contains [1,2].
randomizedSet.getRandom(); // getRandom() should return either 1 or 2 randomly.
randomizedSet.remove(1); // Removes 1 from the set, returns true. Set now contains [2].
randomizedSet.insert(2); // 2 was already in the set, so return false.
randomizedSet.getRandom(); // Since 2 is the only number in the set, getRandom() will always return 2.
Constraints
-2^31 <= val <= 2^31 - 1
- At most
2 * 10^5
calls will be made toinsert
,remove
, andgetRandom
.- There will be at least one element in the data structure when
getRandom
is called.
解题思路
这道题的重点是三种操作都需要达到平均 O(1)
时间,可以有以下判断:
- 随机读取元素,所以存储方式肯定要是数组,这里用 vector
- 常数时间查找元素,所以需要用哈希表存储下标
- 在去除的时候需要在常数时间内去除,并且还要保持数组紧凑(因为要保证 getRamdon 可以常数时间内获得)因此,在去除的时候可以先把元素交换到数组末尾在进行常数时间的pop out
代码如下:
#include <unordered_map>
#include <vector>
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 (num_index.find(val) == num_index.end()) {
nums.push_back(val);
num_index[val] = nums.size() - 1;
return true;
}
return false;
}
/** Removes a value from the set. Returns true if the set contained the
* specified element. */
bool remove(int val) {
if (num_index.find(val) != num_index.end()) {
int index = num_index[val];
num_index[nums[nums.size() - 1]] = index;
std::swap(nums[index], nums[nums.size() - 1]);
nums.pop_back();
num_index.erase(val);
return true;
}
return false;
}
/** Get a random element from the set. */
int getRandom() { return nums[rand() % nums.size()]; }
private:
std::unordered_map<int, int> num_index;
std::vector<int> nums;
};
- 时间复杂度:
O(1)
- 空间复杂度:
O(n)
GitHub 代码同步地址: 380.InsertDeleteGetRandomO1.cpp
其他题目: GitHub: Leetcode-C++-Solution 博客: Leetcode-Solutions