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/*
* @lc app=leetcode.cn id=706 lang=cpp
*
* [706] Design HashMap
*
* https://leetcode-cn.com/problems/design-hashmap/description/
*
* algorithms
* Easy (53.46%)
* Total Accepted: 1.9K
* Total Submissions: 3.5K
* Testcase Example: '["MyHashMap","put","put","get","get","put","get", "remove", "get"]\n[[],[1,1],[2,2],[1],[3],[2,1],[2],[2],[2]]'
*
* 不使用任何内建的哈希表库设计一个哈希映射
*
* 具体地说,你的设计应该包含以下的功能
*
*
* put(key, value):向哈希映射中插入(键,值)的数值对。如果键对应的值已经存在,更新这个值。
* get(key):返回给定的键所对应的值,如果映射中不包含这个键,返回-1。
* remove(key):如果映射中存在这个键,删除这个数值对。
*
*
*
* 示例:
*
*
* MyHashMap hashMap = new MyHashMap();
* hashMap.put(1, 1);          
* hashMap.put(2, 2);        
* hashMap.get(1);            // 返回 1
* hashMap.get(3);            // 返回 -1 (未找到)
* hashMap.put(2, 1);         // 更新已有的值
* hashMap.get(2);            // 返回 1
* hashMap.remove(2);         // 删除键为2的数据
* hashMap.get(2);            // 返回 -1 (未找到)
*
*
*
* 注意:
*
*
* 所有的值都在 [1, 1000000]的范围内。
* 操作的总数目在[1, 10000]范围内。
* 不要使用内建的哈希库。
*
*
*/
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 -1;
}
return data[hashkey][key/1000];
}
/** 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;
};
/**
* 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);
*/
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