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【LeetCode】706. Design HashMap 解题报告(Python)


id: fuxuemingzhu

个人博客: ​​​http://fuxuemingzhu.cn/​​​


目录


  • ​​题目描述​​
  • ​​题目大意​​
  • ​​解题方法​​
  • ​​日期​​


题目地址:​​https://leetcode.com/problems/design-hashmap/description/​​

题目描述

Design a HashMap without using any built-in hash table libraries.

To be specific, your design should include these two functions:


  • ​put(key, value)​​ : Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.
  • ​get(key)​​: Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
  • ​remove(key)​​ : Remove the mapping for the value key if this map contains the mapping for the key.

Example:

MyHashMap hashMap = new MyHashMap();
hashMap.put(1, 1);
hashMap.put(2, 2);
hashMap.get(1); // returns 1
hashMap.get(3); // returns -1 (not found)
hashMap.put(2, 1); // update the existing value
hashMap.get(2); // returns 1
hashMap.remove(2); // remove the mapping for 2
hashMap.get(2); // returns -1 (not found)

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 HashMap library.

题目大意

动手实现一个hashmap.不能用已经内置的函数。

解题方法

这个题和​​705. Design HashSet​​基本一样啊。705是要设计hashset,当时只要把某个位置设置成1,就表示这个元素存在了即可。这个题只需要把当时的设置成1改成设置成对应的value。

写这个题的时候就要考虑,不能把每个位置初始化成0,因为这样会和value值冲突。即加入value就是0,那么这个位置的0不知道怎么判断。所以应该初始化None,后面对这个位置是否存在元素的判断也是必须判断==None而不是not的方式进行判断。

代码如下:

class MyHashMap:

def __init__(self):
"""
Initialize your data structure here.
"""
self.buckets = 1000
self.itemsPerBuckect = 1001
self.hashmap = [[] for _ in range(self.buckets)]

def hash(self, key):
return key % self.buckets

def pos(self, key):
return key // self.buckets

def put(self, key, value):
"""
value will always be positive.
:type key: int
:type value: int
:rtype: void
"""
hashkey = self.hash(key)
if not self.hashmap[hashkey]:
self.hashmap[hashkey] = [None] * self.itemsPerBuckect
self.hashmap[hashkey][self.pos(key)] = value

def get(self, key):
"""
Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key
:type key: int
:rtype: int
"""
hashkey = self.hash(key)
if (not self.hashmap[hashkey]) or self.hashmap[hashkey][self.pos(key)] == None:
return -1
else:
return self.hashmap[hashkey][self.pos(key)]

def remove(self, key):
"""
Removes the mapping of the specified value key if this map contains a mapping for the key
:type key: int
:rtype: void
"""
hashkey = self.hash(key)
if self.hashmap[hashkey]:
self.hashmap[hashkey][self.pos(key)] = None


# Your MyHashMap object will be instantiated and called as such:
# obj = MyHashMap()
# obj.put(key,value)
# param_2 = obj.get(key)
# obj.remove(key)

上面的做法比较省内存,如果不用省内存的话,可以直接开出来这么大的数组即可。事实上,1000000大小的数组,内存不会超的。

另外一个技巧就是,把数组初始化为-1,这样既不会和原始的数据冲突,而且当查询的时候,可以直接返回这个数值,正好是题目的要求。

class MyHashMap(object):

def __init__(self):
"""
Initialize your data structure here.
"""
self.bitmap = [[-1] * 1000 for _ in range(1001)]

def put(self, key, value):
"""
value will always be non-negative.
:type key: int
:type value: int
:rtype: void
"""
row, col = key / 1000, key % 1000
self.bitmap[row][col] = value

def get(self, key):
"""
Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key
:type key: int
:rtype: int
"""
row, col = key / 1000, key % 1000
return self.bitmap[row][col]

def remove(self, key):
"""
Removes the mapping of the specified value key if this map contains a mapping for the key
:type key: int
:rtype: void
"""
row, col = key / 1000, key % 1000
self.bitmap[row][col] = -1


# Your MyHashMap object will be instantiated and called as such:
# obj = MyHashMap()
# obj.put(key,value)
# param_2 = obj.get(key)
# obj.remove(key)

如果直接使用一维数组的话,那么也可以通过,其实更简单了。

class MyHashMap(object):

def __init__(self):
"""
Initialize your data structure here.
"""
self.bitmap = [-1] * 1000001

def put(self, key, value):
"""
value will always be non-negative.
:type key: int
:type value: int
:rtype: void
"""
self.bitmap[key] = value

def get(self, key):
"""
Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key
:type key: int
:rtype: int
"""
return self.bitmap[key]

def remove(self, key):
"""
Removes the mapping of the specified value key if this map contains a mapping for the key
:type key: int
:rtype: void
"""
self.bitmap[key] = -1


# Your MyHashMap object will be instantiated and called as such:
# obj = MyHashMap()
# obj.put(key,value)
# param_2 = obj.get(key)
# obj.remove(key)

日期

2018 年 7 月 12 日 —— 天阴阴地潮潮,已经连着两天这样了

2018 年 11 月 13 日 —— 时间有点快



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