[leetcode] 706. Design HashMap

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

[leetcode] 706. Design HashMap的相关教程结束。