[코테] 리트코드 해시 테이블 706. Design HashMap

706. Design HashMap

문제 링크

https://leetcode.com/problems/design-hashmap/description/

문제 설명

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.

제한 사항

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

입출력 예 #1

1. Input : [“MyHashMap”, “put”, “put”, “get”, “get”, “put”, “get”, “remove”, “get”] [[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
2. Output : [null, null, null, 1, -1, null, 1, null, -1]
3. 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]]

class ListNode:
    def __init__(self, key=None, value=None):
        self.key = key
        self.value = value
        self.next = None

class MyHashMap:

    # 초기화
    def __init__(self):
        self.size = 1000
        self.table = collections.defaultdict(ListNode)

    # 삽입
    def put(self, key: int, value: int) -> None:
        index = key % self.size
        # 인덱스에 노드가 없다면 삽입 후 종료
        if self.table[index].value is None:
            self.table[index] = ListNode(key, value)
            return

        # 인덱스에 노드가 존재하는 경우 연결 리스트 처리
        p = self.table[index]
        while p:
            if p.key == key:
                p.value = value
                return
            if p.next is None:
                break
            p = p.next
        p.next = ListNode(key, value)

    # 조회
    def get(self, key: int) -> int:
        index = key % self.size
        if self.table[index].value is None:
            return -1

        # 노드가 존재할 때 일치하는 키 탐색
        p = self.table[index]
        while p:
            if p.key == key:
                return p.value
            p = p.next
        return -1

    # 삭제
    def remove(self, key: int) -> None:
        index = key % self.size
        if self.table[index].value is None:
            return

        # 인덱스의 첫 번째 노드일 때 삭제 처리
        p = self.table[index]
        if p.key == key:
            self.table[index] = ListNode() if p.next is None else p.next
            return

        # 연결 리스트 노드 삭제
        prev = p
        while p:
            if p.key == key:
                prev.next = p.next
                return
            prev, p = p, p.next