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/* | ||
Design a HashMap without using any built-in hash table libraries. | ||
Designing a hash table using the chaining method, where a vector holds a list of pairs representing the key-value mappings. | ||
Time: O(n) | ||
Space: O(n) | ||
*/ | ||
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class MyHashMap { | ||
public: | ||
const int N = 10010; | ||
vector<list<pair<int, int>>> h; | ||
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/** Initialize your data structure here. */ | ||
MyHashMap() { | ||
h = vector<list<pair<int, int>>>(N); | ||
} | ||
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list<pair<int, int>>::iterator find(int key) { | ||
int t = key % N; | ||
for (auto it = h[t].begin(); it != h[t].end(); it ++ ) { | ||
if (it->first == key) | ||
return it; | ||
} | ||
return h[t].end(); | ||
} | ||
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/** value will always be non-negative. */ | ||
void put(int key, int value) { | ||
auto t = key % N; | ||
auto it = find(key); | ||
if (it == h[t].end()) h[t].push_back({key, value}); | ||
else it->second = value; | ||
} | ||
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/** 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) { | ||
auto t = key % N; | ||
auto it = find(key); | ||
if (it != h[t].end()) return it->second; | ||
return -1; | ||
} | ||
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/** Removes the mapping of the specified value key if this map contains a mapping for the key */ | ||
void remove(int key) { | ||
auto t = key % N; | ||
auto it = find(key); | ||
if (it != h[t].end()) h[t].erase(it); | ||
} | ||
}; | ||
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/** | ||
* 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); | ||
*/ |