题目
Design 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
.
思路
关键在于采用高效的散列函数。
代码
python版本:
# Runtime: 9894 ms, faster than 5.00% of Python3 online submissions for Design HashMap.
class MyHashMap:
def __init__(self):
self.hashTable = [[]]*10000
def put(self, key: int, value: int) -> None:
hash = key % 10000
for i in range(len(self.hashTable[hash])):
if self.hashTable[hash][i][0] == key:
self.hashTable[hash][i] = (key, value)
return
self.hashTable[hash].append((key, value))
def get(self, key: int) -> int:
hash = key % 10000
for i in range(len(self.hashTable[hash])):
if self.hashTable[hash][i][0] == key:
return self.hashTable[hash][i][1]
return -1
def remove(self, key: int) -> None:
hash = key % 10000
for i in range(len(self.hashTable[hash])):
if self.hashTable[hash][i][0] == key:
self.hashTable[hash].pop(i)
return