LeetCode 706 Design HashMap

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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:

<pre>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]] </pre>

Constraints:

0 <= key, value <= 10<sup>6</sup>
At most 10<sup>4</sup> calls will be made to put, get, and remove.

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题目大意：

设计HashMap

LL解题思路(推荐)：

大学学到的方法，用Array实现，将key mod prime num找到index插入。难点在于冲突处理，这里用chaining方法，也就是LL

解题步骤：

N/A

注意事项：

迭代LL时候，每种方法put, get, remove都不同。put只能迭代到最后一个不能到None，因为要从尾部加入。 get正常一个个迭代。remove要从parent也就是it.next迭代因为要删除节点。

Python代码：

class MyHashMap(TestCases):

    def __init__(self):
        self.key_space = 997
        self.buckets = [ListNode(-1, -1) for _ in range(self.key_space)]

    def put(self, key: int, value: int) -> None:
        index = key % self.key_space
        it = self.buckets[index]
        while it:
            if it.key == key:
                it.val = value
                return
            if not it.next:
                break
            it = it.next
        it.next = ListNode(key, value)

    def get(self, key: int) -> int:
        index = key % self.key_space
        it = self.buckets[index]
        while it:
            if it.key == key:
                return it.val
            it = it.next
        return -1

    def remove(self, key: int) -> None:
        index = key % self.key_space
        it = self.buckets[index]
        while it.next:
            if it.next.key == key:
                tmp = it.next
                it.next, tmp.next = it.next.next, None
                return
            it = it.next


class ListNode:

    def __init__(self, key, val):
        self.key = key
        self.val = val
        self.next = None

算法分析：

时间复杂度为O(k)，空间复杂度O(n), k为冲突数

数组算法II解题思路：

较容易实现，remove复杂度稍差，但是最差情况也是同上

Python代码：

class MyHashMap(TestCases):

    def __init__(self):
        self.key_space = 997
        self.buckets = [[] for _ in range(self.key_space)]

    def put(self, key: int, value: int) -> None:
        index = key % self.key_space
        for i, [_key, _val] in enumerate(self.buckets[index]):
            if _key == key:
                self.buckets[index][i] = [key, value]
                return
        self.buckets[index].append([key, value])

    def get(self, key: int) -> int:
        index = key % self.key_space
        for _key, _val in self.buckets[index]:
            if _key == key:
                return _val
        return -1

    def remove(self, key: int) -> None:
        index = key % self.key_space
        for i, [_key, _val] in enumerate(self.buckets[index]):
            if _key == key:
                self.buckets[index].pop(i)
                return

算法分析：

时间复杂度为O(k)，空间复杂度O(n), k为冲突数