[LeetCode] Design HashMap 設計HashMap

 

Design a HashMap without using any built-in hash table libraries.

To be specific, your design should include these functions:

• put(key, value) : Insert a (key, value) pair into the HashMap. If the value already exists in the HashMap, update the value.
• get(key): Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
• remove(key) : Remove the mapping for the value key if this map contains the mapping for the key.

Example:

MyHashMap hashMap = new MyHashMap();
hashMap.put(1, 1);          
hashMap.put(2, 2);         
hashMap.get(1);            // returns 1
hashMap.get(3);            // returns -1 (not found)
hashMap.put(2, 1);          // update the existing value
hashMap.get(2);            // returns 1 
hashMap.remove(2);          // remove the mapping for 2
hashMap.get(2);            // returns -1 (not found) 

Note:

• All keys and values will be in the range of [0, 1000000].
• The number of operations will be in the range of [1, 10000].
• Please do not use the built-in HashMap library.

 

這道題讓咱們設計一個HashMap的數據結構，不能使用自帶的哈希表，跟以前那道Design HashSet很相似，以前那道的兩種解法在這裏也是行得通的，既然題目中說了數字的範圍不會超過1000000，那麼咱們就申請這麼大空間的數組，只需將數組的初始化值改成-1便可。空間足夠大了，咱們就能夠直接創建映射，移除時就將映射值重置爲-1，因爲默認值是-1，因此獲取映射值就能夠直接去，參見代碼以下：

 

解法一：

class MyHashMap {
public:
    /** Initialize your data structure here. */
    MyHashMap() {
        data.resize(1000000, -1);
    }
    
    /** value will always be non-negative. */
    void put(int key, int value) {
        data[key] = value;
    }
    
    /** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
    int get(int key) {
        return data[key];
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    void remove(int key) {
        data[key] = -1;
    }

private:
    vector<int> data;
};

 

咱們能夠來適度的優化一下空間複雜度，因爲存入HashMap的映射對兒也許不會跨度很大，那麼直接就申請長度爲1000000的數組可能會有些浪費，那麼咱們其實可使用1000個長度爲1000的數組來代替，那麼就要用個二維數組啦，實際上開始咱們只申請了1000個空數組，對於每一個要處理的元素，咱們首先對1000取餘，獲得的值就看成哈希值，對應咱們申請的那1000個空數組的位置，在創建映射時，一旦計算出了哈希值，咱們將對應的空數組resize爲長度1000，而後根據哈希值和key/1000來肯定具體的加入映射值的位置。獲取映射值時，計算出哈希值，若對應的數組不爲空，直接返回對應的位置上的值。移除映射值同樣的，先計算出哈希值，若是對應的數組不爲空的話，找到對應的位置並重置爲-1。參見代碼以下：

 

解法二：

class MyHashMap {
public:
    /** Initialize your data structure here. */
    MyHashMap() {
        data.resize(1000, vector<int>());
    }
    
    /** value will always be non-negative. */
    void put(int key, int value) {
        int hashKey = key % 1000;
        if (data[hashKey].empty()) {
            data[hashKey].resize(1000, -1);
        } 
        data[hashKey][key / 1000] = value;
    }
    
    /** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
    int get(int key) {
        int hashKey = key % 1000;
        if (!data[hashKey].empty()) {
            return data[hashKey][key / 1000];
        } 
        return -1;
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    void remove(int key) {
        int hashKey = key % 1000;
        if (!data[hashKey].empty()) {
            data[hashKey][key / 1000] = -1;
        } 
    }

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

 

相似題目：

Design HashSet

 

參考資料：

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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