leetcode706 - Design HashMap - easy

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.

 

最简单的办法，用一个1000000大小的array，idx是key，idx上的值是value。但一开始key不会都分散的那么开，所以用一个hash function。

定义hash function为f(x) = x%(some prime number) - so keys are uniformly distributed.

How to handle collision: using buckets for the same hash output (or address).

比如说大家都mode 997, 那1和998会被mapped到同一个位置，但注意这些数相互之前肯定都差997，所以key/997就能将他们区分开。

每当有key要映射到一个addr的时候，把这个addr展开成一排buckets，相当于准备做chaining。

 

实现：Time O(n/k), k-hashNum we chose

class MyHashMap {
    
    int hashNum;
    vector<vector<int>> data;
    
public:
    /** Initialize your data structure here. */
    MyHashMap() {
        hashNum = 1031;
        data.resize(hashNum, vector<int>());
    }
    
    /** value will always be non-negative. */
    void put(int key, int value) {
        int hashRes = key%hashNum;
        if (data[hashRes].empty())
            data[hashRes].resize(970, -1);
        data[hashRes][key/hashNum] = 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 hashRes = key%hashNum;
        if (data[hashRes].empty())
            return -1;
        return data[hashRes][key/hashNum] >=0 ? data[hashRes][key/hashNum] : -1 ;
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    void remove(int key) {
        int hashRes = key%hashNum;
        if (!data[hashRes].empty())
            data[hashRes][key/hashNum] = -1;
    }
};