我想在Python中实现一个HashMap。我想让用户输入。根据他的输入，我正在从HashMap中检索一些信息。如果用户输入HashMap的一个键，我想检索相应的值。
我如何在Python中实现这个功能?
HashMap<String,String> streetno=new HashMap<String,String>();
streetno.put("1", "Sachin Tendulkar");
streetno.put("2", "Dravid");
streetno.put("3","Sehwag");
streetno.put("4","Laxman");
streetno.put("5","Kohli")
当前回答
class HashMap:
def __init__(self):
self.size = 64
self.map = [None] * self.size
def _get_hash(self, key):
hash = 0
for char in str(key):
hash += ord(char)
return hash % self.size
def add(self, key, value):
key_hash = self._get_hash(key)
key_value = [key, value]
if self.map[key_hash] is None:
self.map[key_hash] = list([key_value])
return True
else:
for pair in self.map[key_hash]:
if pair[0] == key:
pair[1] = value
return True
else:
self.map[key_hash].append(list([key_value]))
return True
def get(self, key):
key_hash = self._get_hash(key)
if self.map[key_hash] is not None:
for pair in self.map[key_hash]:
if pair[0] == key:
return pair[1]
return None
def delete(self, key):
key_hash = self._get_hash(key)
if self.map[key_hash] is None :
return False
for i in range(0, len(self.map[key_hash])):
if self.map[key_hash][i][0] == key:
self.map[key_hash].pop(i)
return True
def print(self):
print('---Phonebook---')
for item in self.map:
if item is not None:
print(str(item))
h = HashMap()
其他回答
Python字典是一种内置类型，支持键值对。它是相对于Java的HashMap最近的内置数据结构。
你可以声明一个键值对为值的dict:
streetno = {
"1": "Sachin Tendulkar",
"2": "Dravid",
"3": "Sehwag",
"4": "Laxman",
"5": "Kohli"
}
你也可以在创建后设置一个键值映射:
streetno = {}
streetno["1"] = "Sachin Tendulkar"
print(streetno["1"]) # => "Sachin Tendulkar"
另一种创建字典的方法是使用内置函数dict()，但这仅在键是有效标识符时有效:
streetno = dict(one="Sachin Tendulkar", two="Dravid")
print(streetno["one"]) # => "Sachin Tendulkar"
Python Counter在这种情况下也是一个很好的选择:
from collections import Counter
counter = Counter(["Sachin Tendulkar", "Sachin Tendulkar", "other things"])
print(counter)
这将返回一个包含列表中每个元素计数的dict:
Counter({'Sachin Tendulkar': 2, 'other things': 1})
在python中，你会使用字典。
它是python中非常重要的类型，经常使用。
您可以轻松地创建一个
name = {}
字典有很多方法:
# add entries:
>>> name['first'] = 'John'
>>> name['second'] = 'Doe'
>>> name
{'first': 'John', 'second': 'Doe'}
# you can store all objects and datatypes as value in a dictionary
# as key you can use all objects and datatypes that are hashable
>>> name['list'] = ['list', 'inside', 'dict']
>>> name[1] = 1
>>> name
{'first': 'John', 'second': 'Doe', 1: 1, 'list': ['list', 'inside', 'dict']}
你不能影响字典的顺序。
哈希映射是Python中内置的，它们被称为字典:
streetno = {} #create a dictionary called streetno
streetno["1"] = "Sachin Tendulkar" #assign value to key "1"
用法:
"1" in streetno #check if key "1" is in streetno
streetno["1"] #get the value from key "1"
有关更多信息，如内置方法等，请参阅文档。它们很棒，在Python程序中非常常见(这并不奇怪)。
这就是我对LeetCode问题706的解决方案:
一个哈希映射类，有三个方法:get、put和remove
class Item:
def __init__(self, key, value):
self.key = key
self.value = value
self.next = None
class MyHashMap:
def __init__(self, size=100):
self.items = [None] * size
self.size = size
def _get_index(self, key):
return hash(key) & self.size-1
def put(self, key: int, value: int) -> None:
index = self._get_index(key)
item = self.items[index]
if item is None:
self.items[index] = Item(key, value)
else:
if item.key == key:
item.value = value
else:
while True:
if item.key == key:
item.value = value
return
else:
if not item.next:
item.next = Item(key, value)
return
item = item.next
def get(self, key: int) -> int:
index = self._get_index(key)
if self.items[index] is None:
return -1
item = self.items[index]
while True:
if item.key == key:
return item.value
else:
if item.next:
item = item.next
else:
return -1
def remove(self, key: int) -> None:
value = self.get(key)
if value > -1:
index = self._get_index(key)
item = self.items[index]
if item.key == key:
self.items[index] = item.next if item.next else None
return
while True:
if item.next and item.next.key == key:
item.next = item.next.next
return
else:
if item.next:
item = item.next
else:
return
