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 1 is the only number in the set, getRandom always return 1.
randomSet.getRandom();

设计一个数据结构在O(1)时间内完成:insert(val),remove(val),getRandom()

由于是O(1)时间,要用哈希表 + 数组(HashMap + Array),数组用来保存数字,哈希表用来建立每个数字和其在数组中的位置之间的映射。

插入操作:先判断是否在HashMap里,如果存在则不需要插入,直接返回false。不存在的话,把此数插入到数组的末尾,然后在HashMap中建立数字和其位置的映射。

删除操作:先判断是否在HashMap里,如果没有,直接返回false。由于HashMap的删除是O(1)时间,而数组并不是。为了保证了O(1)时间内的删除,把要删除的数字和数组的最后一个数调换个位置,然后修改对应的HashMap中的值,删除数组的最后一个元素即可。

返回随机数:随机生成一个数组位置,返回该位置上的数字。

Java:

public class RandomizedSet {
Map<Integer, Integer> map;//<val, index in arrlist>
List<Integer> list;//keep a record of value for get random
int size;//map and list are the same size
Random rand;
/** Initialize your data structure here. */
public RandomizedSet() {
this.map = new HashMap<>();
this.list = new ArrayList<>();
this.size = 0;
this.rand = new Random();
} /** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
public boolean insert(int val) {
if(map.containsKey(val)) return false;
map.put(val, size);
list.add(val);
size++;
return true;
} /** Removes a value from the set. Returns true if the set contained the specified element. */
public boolean remove(int val) {
if(map.containsKey(val)){
int last = list.get(size - 1);
list.set(map.get(val), last);//use last added ele to fill the element to be removed
map.put(last, map.get(val));//update last added ele's index in list
map.remove(val);
list.remove(size - 1);//remove at end is constant
size--;
return true;
}
return false;
} /** Get a random element from the set. */
public int getRandom() {
return list.get(rand.nextInt(size));
}
}

Python:

import random
class RandomizedSet(object): def __init__(self):
"""
Initialize your data structure here.
"""
self.dataMap = {}
self.dataList = [] def insert(self, val):
"""
Inserts a value to the set. Returns true if the set did not already contain the specified element.
:type val: int
:rtype: bool
"""
if val in self.dataMap:
return False
self.dataMap[val] = len(self.dataList)
self.dataList.append(val)
return True def remove(self, val):
"""
Removes a value from the set. Returns true if the set contained the specified element.
:type val: int
:rtype: bool
"""
if val not in self.dataMap:
return False
idx = self.dataMap[val]
tail = self.dataList.pop()
if idx < len(self.dataList):
self.dataList[idx] = tail
self.dataMap[tail] = idx
del self.dataMap[val]
return True def getRandom(self):
"""
Get a random element from the set.
:rtype: int
"""
return random.choice(self.dataList)  

C++:

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 (m.count(val)) return false;
nums.push_back(val);
m[val] = nums.size() - 1;
return true;
} /** Removes a value from the set. Returns true if the set contained the specified element. */
bool remove(int val) {
if (!m.count(val)) return false;
int last = nums.back();
m[last] = m[val];
nums[m[val]] = last;
nums.pop_back();
m.erase(val);
return true;
} /** Get a random element from the set. */
int getRandom() {
return nums[rand() % nums.size()];
}
private:
vector<int> nums;
unordered_map<int, int> m;
};

类似题目:

[LeetCode] 381. Insert Delete GetRandom O(1) - Duplicates allowed 插入删除和获得随机数O(1)时间 - 允许重复  

All LeetCode Questions List 题目汇总

[LeetCode] 380. Insert Delete GetRandom O(1) 插入删除获得随机数O(1)时间的更多相关文章

  1. LeetCode 380. Insert Delete GetRandom O(1)

    380. Insert Delete GetRandom O(1) Add to List Description Submission Solutions Total Accepted: 21771 ...

  2. leetcode 380. Insert Delete GetRandom O(1) 、381. Insert Delete GetRandom O(1) - Duplicates allowed

    380. Insert Delete GetRandom O(1) 实现插入.删除.获得随机数功能,且时间复杂度都在O(1).实际上在插入.删除两个功能中都包含了查找功能,当然查找也必须是O(1). ...

  3. [LeetCode] 380. Insert Delete GetRandom O(1) 常数时间内插入删除和获得随机数

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

  4. LeetCode 380. Insert Delete GetRandom O(1) (插入删除和获得随机数 常数时间)

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

  5. [leetcode]380. Insert Delete GetRandom O(1)常数时间插入删除取随机值

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

  6. LeetCode 380. Insert Delete GetRandom O(1) 常数时间插入、删除和获取随机元素(C++/Java)

    题目: Design a data structure that supports all following operations in averageO(1) time. insert(val): ...

  7. [leetcode]380. Insert Delete GetRandom O(1)设计数据结构,实现存,删,随机取的时间复杂度为O(1)

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

  8. [LeetCode] 381. Insert Delete GetRandom O(1) - Duplicates allowed 插入删除和获得随机数O(1)时间 - 允许重复

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

  9. LeetCode 381. Insert Delete GetRandom O(1) - Duplicates allowed O(1) 时间插入、删除和获取随机元素 - 允许重复(C++/Java)

    题目: Design a data structure that supports all following operations in averageO(1) time. Note: Duplic ...

随机推荐

  1. Kotlin函数式编程范式深入剖析

    继续学习Kotlin的函数式编程,先定义一个高阶函数: 其实上面这种调用方式在Kotlin用得不多,反而是将Lambda表达式放到方法体中使用得较频繁,如下: 接下来定义一个扩展方法,用来对字符串进行 ...

  2. requireJS的基本使用

    requireJS的基本使用 一.总结 一句话总结: requireJS是js端模块化开发,主要是实现js的异步加载,和管理模块之间的依赖关系,便于代码的编写和维护 1.页面加载的js文件过多的缺点是 ...

  3. 1220 Vue与Django前后端的结合

    目录 vue的安装 Vue前端的设置 页面的分布 后台数据的替换 css样式 Django的配置 国际化配置 axios插件安装 CORS跨域问题(同源策略) 处理跨域问题: cors插件 axios ...

  4. python_并发编程——数据共享

    1.数据共享 实现进程之间的数据共享 from multiprocessing import Manager,Process class MyPro(Process): def __init__(se ...

  5. Non-boring sequences(启发式分治)

    题意:一个序列被称作是不无聊的,当且仅当,任意一个连续子区间,存在一个数字只出现了一次,问给定序列是否是不无聊的. 思路:每次找到一个只出现了一次的点,其位置的pos,那么继续分治[L,pos-1], ...

  6. 基于思岚A1激光雷达+OpenGL+VS2017的Ramer-Douglas-Peucker算法的实现

    时隔两年 又借到了之前的那个激光雷达,最老版本的思岚A1,甚至不支持新的固件,并且转接板也不见了,看了下淘宝店卖¥80,但是官方提供了一个基于STM32的实现方式,于是我估摸着这个转接板只是一个普通的 ...

  7. 004——转载—Word2016“此功能看似已中断 并需要修复”问题解决办法

    解决办法如下: 在Win10系统上安装 Office 2016 之后,每次打开Word文档可能都会提示“很抱歉,此功能看似已中断,并需要修复,请使用Windows 控制面板中的“程序和功能”选项修复M ...

  8. 003-转载-keil-STM32硬件错误HardFault_Handler的处理方法

    (一)参考文献:https://blog.csdn.net/electrocrazy/article/details/78173558 在用Keil对STM32的程序进行仿真时程序有时会跑飞,停止仿真 ...

  9. 洛谷 P1821 [USACO07FEB]银牛派对Silver Cow Party 题解

    P1821 [USACO07FEB]银牛派对Silver Cow Party 题目描述 One cow from each of N farms (1 ≤ N ≤ 1000) conveniently ...

  10. 记录一次SpringBoot实现AOP编程

    需求 最近碰到一个问题,需要对关键操作的入参和返回值进行记录,并不是使用log记录,而是插入到数据库中. 思路:如果采用硬编码,在每个操作后都添加,会产生大量重复代码.因而打算使用自定义注解,通过AO ...