706. Design HashMap
Category: /leetcodeDesign a HashMap without using any built-in hash table libraries.
Implement the MyHashMap class:
- MyHashMap() initializes the object with an empty map.
- void put(int key, int value) inserts a (key, value) pair into the HashMap. If the key already exists in the map, update the corresponding value.
- int get(int key) returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
- void remove(key) removes the key and its corresponding value if the map contains the mapping for the key.
Example 1:
Input
["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
Output
[null, null, null, 1, -1, null, 1, null, -1]
Explanation
MyHashMap myHashMap = new MyHashMap();
myHashMap.put(1, 1); // The map is now [[1,1]]
myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
myHashMap.get(1); // return 1, The map is now [[1,1], [2,2]]
myHashMap.get(3); // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
myHashMap.get(2); // return 1, The map is now [[1,1], [2,1]]
myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
myHashMap.get(2); // return -1 (i.e., not found), The map is now [[1,1]]
Constraints:
0 <= key, value <= 106
At most 104 calls will be made to put, get, and remove.
Solution
def __init__(self):
"""
Initialize your data structure here.
"""
self.cnt = 1000
self.bucket = [None] * self.cnt
def put(self, key: int, value: int) -> None:
"""
value will always be non-negative.
"""
idx = key % self.cnt
item = self.bucket[idx]
if item is None:
self.bucket[idx] = [[key, value]]
else:
for i in item:
if key == i[0]:
i[1] = value
return
item.append([key, value])
def get(self, key: int) -> int:
"""
Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key
"""
idx = key % self.cnt
item = self.bucket[idx]
if item:
for k, v in item:
if k == key: return v
return -1
def remove(self, key: int) -> None:
"""
Removes the mapping of the specified value key if this map contains a mapping for the key
"""
idx = key % self.cnt
item = self.bucket[idx]
if item:
for pos, i in enumerate(item):
if i[0] == key:
del item[pos]
return
