[LeetCode] 706. Design HashMap #706

grandyang · 2019-05-30T16:25:35Z

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

这道题让我们设计一个 HashMap 的数据结构，不能使用自带的哈希表，跟之前那道 Design HashSet 很类似，之前那道的两种解法在这里也是行得通的，既然题目中说了数字的范围不会超过 1000000，那么就申请这么大空间的数组，只需将数组的初始化值改为 -1 即可。空间足够大了，就可以直接建立映射，移除时就将映射值重置为 -1，由于默认值是 -1，所以获取映射值就可以直接去，参见代码如下：

解法一：

class MyHashMap {
public:
    MyHashMap() {
        data.resize(1e6 + 1, -1);
    }
    void put(int key, int value) {
        data[key] = value;
    }
    int get(int key) {
        return data[key];
    }
    void remove(int key) {
        data[key] = -1;
    }

private:
    vector<int> data;
};

我们可以来适度的优化一下空间复杂度，由于存入 HashMap 的映射对儿也许不会跨度很大，那么直接就申请长度为 1000000 的数组可能会有些浪费，其实可以使用 1000 个长度为 1000 的数组来代替，那么就要用个二维数组啦，实际上开始只申请了 1000 个空数组，对于每个要处理的元素，首先对 1000 取余，得到的值就当作哈希值，对应申请的那 1000 个空数组的位置，在建立映射时，一旦计算出了哈希值，将对应的空数组 resize 为长度 1000，然后根据哈希值和 key/1000 来确定具体的加入映射值的位置。获取映射值时，计算出哈希值，若对应的数组不为空，直接返回对应的位置上的值。移除映射值一样的，先计算出哈希值，如果对应的数组不为空的话，找到对应的位置并重置为 -1。参见代码如下：

解法二：

class MyHashMap {
public:
    MyHashMap() {
        data.resize(1001, vector<int>());
    }
    void put(int key, int value) {
        int hashKey = key % 1000;
        if (data[hashKey].empty()) {
            data[hashKey].resize(1001, -1);
        } 
        data[hashKey][key / 1000] = value;
    }
    int get(int key) {
        int hashKey = key % 1000;
        if (!data[hashKey].empty()) {
            return data[hashKey][key / 1000];
        } 
        return -1;
    }
    void remove(int key) {
        int hashKey = key % 1000;
        if (!data[hashKey].empty()) {
            data[hashKey][key / 1000] = -1;
        } 
    }

private:
    vector<vector<int>> data;
};

Github 同步地址：

#706

类似题目：

Design HashSet

参考资料：

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

https://leetcode.com/problems/design-hashmap/discuss/152746/Java-Solution

https://leetcode.com/problems/design-hashmap/discuss/184764/Easy-C%2B%2B-Solution-beats-98.01(52-msec)-using-array-of-vectors

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