【LeetCode】380. Insert Delete GetRandom O(1) 解题报告(Python)
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址:https://leetcode.com/problems/insert-delete-getrandom-o1/description/
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.
- 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,有三个方法:插入、删除、随机选取一个数值。要求平均的时间复杂度是O(1).
解题方法
插入删除的时间复杂度要求O(1)的话,很容易想起来是set。所以我就用set来实现了。但是随机选取的时候,由于set不能使用索引,所以我先把它转成了list,然后使用随机数来进行索引。不知道python中set转list的时间复杂度是多少,估计最坏情况应该是O(n),这一步没有满足题目的要求,但是也过了。
这个题目没有说清楚如果数据结构为空的时候使用getRandom()应该怎么返回,我觉得是个bug。当然测试用例避开了这一点。
代码如下:
class RandomizedSet(object):
def __init__(self):
"""
Initialize your data structure here.
"""
self.set = set()
self.size = 0
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 not in self.set:
self.set.add(val)
self.size += 1
return True
return False
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 in self.set:
self.set.remove(val)
self.size -= 1
return True
return False
def getRandom(self):
"""
Get a random element from the set.
:rtype: int
"""
ind = random.randint(0, self.size - 1)
return list(self.set)[ind]
# Your RandomizedSet object will be instantiated and called as such:
# obj = RandomizedSet()
# param_1 = obj.insert(val)
# param_2 = obj.remove(val)
# param_3 = obj.getRandom()
参考了一下,发现可以使用字典保存每个元素出现的位置,那么和list结合之后,每次移除一个元素的方式是把list结尾元素对要被移除元素出现的位置进行原地替换,这样就能把时间复杂度降下来。
如果list删除某个位置的元素,那么时间复杂度是O(N),但是如果用最后的元素对该位置进行替换,并且移除最后的元素,时间复杂度能降到O(1)。
特别注意骚操作都在remove里面的,注意位置替换,以及别忘记把list和dict中要移除的元素删除。
class RandomizedSet(object):
def __init__(self):
"""
Initialize your data structure here.
"""
self.nums, self.pos = list(), dict()
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 not in self.pos:
self.nums.append(val)
self.pos[val] = len(self.nums) - 1
return True
return False
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 in self.pos:
idx, last = self.pos[val], self.nums[-1]
self.nums[idx] = last
self.pos[last] = idx
self.nums.pop()
self.pos.pop(val, 0)
return True
return False
def getRandom(self):
"""
Get a random element from the set.
:rtype: int
"""
idx = random.randint(0, len(self.nums) - 1)
return self.nums[idx]
# Your RandomizedSet object will be instantiated and called as such:
# obj = RandomizedSet()
# param_1 = obj.insert(val)
# param_2 = obj.remove(val)
# param_3 = obj.getRandom()
参考资料:
https://leetcode.com/problems/insert-delete-getrandom-o1/discuss/85397/Simple-solution-in-Python
日期
2018 年 9 月 17 日 —— 早上很凉,夜里更凉