leetcode706.DesignHashMap（

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描述

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]]
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Note:

0 <= key, value <= 106
At most 104 calls will be made to put, get, and remove.
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解析

根据题意，就是实现题目中所描述的 hashmap 相关的操作，借助 dict 的相关操作可以完成，不再赘述，这种解法虽然简单，但是有使用内置函数的嫌疑。

解答

 class MyHashMap(object):

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.d = {}

    def put(self, key, value):
        """
        value will always be non-negative.
        :type key: int
        :type value: int
        :rtype: None
        """
        self.d[key] = value

    def get(self, key):
        """
        Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key
        :type key: int
        :rtype: int
        """
        if key in self.d:
            return self.d[key]
        return -1

    def remove(self, key):
        """
        Removes the mapping of the specified value key if this map contains a mapping for the key
        :type key: int
        :rtype: None
        """
        if key in self.d:
            self.d.pop(key)           	      
		
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运行结果

Runtime: 200 ms, faster than 88.65% of Python online submissions for Design HashMap.
Memory Usage: 16.3 MB, less than 95.14% of Python online submissions for Design HashMap.
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解析

另外，可以借用列表来完成相关的操作，将列表中的某个索引当作 key ，将值当作 value ，然后进行相关的操作即可。因为题目中限制了 key 的大小，所以初始化的时候将列表长度设置为 1000001 。

解答

class MyHashMap(object):

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.data = [None] * 1000001

    def put(self, key, value):
        """
        value will always be non-negative.
        :type key: int
        :type value: int
        :rtype: None
        """
        self.data[key] = value

    def get(self, key):
        """
        Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key
        :type key: int
        :rtype: int
        """
        value = self.data[key]
        return value if value != None else -1

    def remove(self, key):
        """
        Removes the mapping of the specified value key if this map contains a mapping for the key
        :type key: int
        :rtype: None
        """
        self.data[key] = None
        


# Your MyHashMap object will be instantiated and called as such:
# obj = MyHashMap()
# obj.put(key,value)
# param_2 = obj.get(key)
# obj.remove(key)
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运行结果

Runtime: 368 ms, faster than 38.96% of Python online submissions for Design HashMap.
Memory Usage: 38 MB, less than 12.01% of Python online submissions for Design HashMap.
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解析

还有一种是使用 Hash 原理来设计 HashMap ，这种比较复杂用到了链表的原理，这个我没有想到，想进一步了解的可以查看官方解法。

原题链接：leetcode.com/problems/de…

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