705. Design HashSet
Read the full problem statement on LeetCode.
Difficulty: easy Acceptance: 67% Topics: Array, Hash Table, Linked List, Design, Hash Function
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Reference solution (spoiler · python)
# Time: O(1)
# Space: O(n)
class ListNode(object):
def __init__(self, key, val):
self.val = val
self.key = key
self.next = None
self.prev = None
class LinkedList(object):
def __init__(self):
self.head = None
self.tail = None
def insert(self, node):
node.next, node.prev = None, None # avoid dirty node
if self.head is None:
self.head = node
else:
self.tail.next = node
node.prev = self.tail
self.tail = node
def delete(self, node):
if node.prev:
node.prev.next = node.next
else:
self.head = node.next
if node.next:
node.next.prev = node.prev
else:
self.tail = node.prev
node.next, node.prev = None, None # make node clean
def find(self, key):
curr = self.head
while curr:
if curr.key == key:
break
curr = curr.next
return curr
class MyHashSet(object):
def __init__(self):
"""
Initialize your data structure here.
"""
self.__data = [LinkedList() for _ in xrange(10000)]
def add(self, key):
"""
:type key: int
:rtype: void
"""
l = self.__data[key % len(self.__data)]
node = l.find(key)
if not node:
l.insert(ListNode(key, 0))
def remove(self, key):
"""
:type key: int
:rtype: void
"""
l = self.__data[key % len(self.__data)]
node = l.find(key)
if node:
l.delete(node)
def contains(self, key):
"""
Returns true if this set did not already contain the specified element
:type key: int
:rtype: bool
"""
l = self.__data[key % len(self.__data)]
node = l.find(key)
return node is not None
Solution from kamyu104/LeetCode-Solutions · MIT
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