Problem

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.

Constraints:

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

Example

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]]

Solution

const int N = 19997;
typedef pair<int, int> PII;

class MyHashMap {
public:
    vector<PII> h[N];

    /** Initialize your data structure here. */
    MyHashMap() {

    }

    int find(vector<PII>& h, int key) {
        for (int i = 0; i < h.size(); i ++ )
            if (h[i].first == key)
                return i;
        return -1;
    }

    /** value will always be non-negative. */
    void put(int key, int value) {
        int t = key % N;
        int k = find(h[t], key);
        if (k == -1) h[t].push_back({key, value});
        else h[t][k].second = 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 t = key % N;
        int k = find(h[t], key);
        if (k == -1) return -1;
        return h[t][k].second;
    }

    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    void remove(int key) {
        int t = key % N;
        int k = find(h[t], key);
        if (k != -1) h[t].erase(h[t].begin() + k);
    }
};

/**
 * Your MyHashMap object will be instantiated and called as such:
 * MyHashMap* obj = new MyHashMap();
 * obj->put(key,value);
 * int param_2 = obj->get(key);
 * obj->remove(key);
 */

标签：2021.03,HashMap,map,int,get,value,key,put,Leetcode
来源： https://blog.csdn.net/sjt091110317/article/details/114798157