题目地址：https://leetcode.com/problems/design-hashset/description/
题目描述
Design a HashSet without using any built-in hash table libraries.
Tobe specific, your design should include these two functions:
- add(value): Insert a value into the HashSet.
- contains(value) : Return whether the value exists in the HashSet or not.
- remove(value): Remove a value in the HashSet. If the value does not exist in the HashSet, do nothing.
Example:
MyHashSet hashSet = new MyHashSet();
hashSet.add(1);
hashSet.add(2);
hashSet.contains(1); // returns true
hashSet.contains(3); // returns false (not found)
hashSet.add(2);
hashSet.contains(2); // returns true
hashSet.remove(2);
hashSet.contains(2); // returns false (already removed)
Note:
- All values will be in the range of [1, 1000000].
- The number of operations will be in the range of [1, 10000].
- Please do not use the built-in HashSet library.
题目大意
动手实现一个hashset.不能用已经内置的函数。
解题方法
位图法
那么直接想到能不能真正模拟hashset呢？通过计算hash，或者一个元素一个坑的方法进行模拟？
参考了一下别人的代码，比我想的要机智一点。这个思路的方法是第一个维度只做hash，第二个维度保存具体元素。这个思想类似于HashMap中的bucket+链表桶。
我觉得这个方法最大的优点就是省内存，因为这种设计当需要的时候，才会产生第二个维度的数据。
代码如下：
class MyHashSet:
def __init__(self):
"""
Initialize your data structure here.
"""
self.buckets = 1000
self.itemsPerBucket = 1001
self.table = [[] for _ in range(self.buckets)]
def hash(self, key):
return key % self.buckets
def pos(self, key):
return key // self.buckets
def add(self, key):
"""
:type key: int
:rtype: void
"""
hashkey = self.hash(key)
if not self.table[hashkey]:
self.table[hashkey] = [0] * self.itemsPerBucket
self.table[hashkey][self.pos(key)] = 1
def remove(self, key):
"""
:type key: int
:rtype: void
"""
hashkey = self.hash(key)
if self.table[hashkey]:
self.table[hashkey][self.pos(key)] = 0
def contains(self, key):
"""
Returns true if this set did not already contain the specified element
:type key: int
:rtype: bool
"""
hashkey = self.hash(key)
return (self.table[hashkey] != []) and (self.table[hashkey][self.pos(key)] == 1)
# Your MyHashSet object will be instantiated and called as such:
# obj = MyHashSet()
# obj.add(key)
# obj.remove(key)
# param_3 = obj.contains(key)
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数组法
直接开辟这么大的一个数组，然后全部设置成False，哪里有数字哪里就是True。空间没有超。
class MyHashSet:
def __init__(self):
"""
Initialize your data structure here.
"""
self.set = [False] * 1000001
def add(self, key):
"""
:type key: int
:rtype: void
"""
self.set[key] = True
def remove(self, key):
"""
:type key: int
:rtype: void
"""
self.set[key] = False
def contains(self, key):
"""
Returns true if this set contains the specified element
:type key: int
:rtype: bool
"""
return self.set[key]
# Your MyHashSet object will be instantiated and called as such:
# obj = MyHashSet()
# obj.add(key)
# obj.remove(key)
# param_3 = obj.contains(key)
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