LintCode-520. Consistent Hashing II
这题要注意的是用一个map< int, int >来存储id和machine之间的映射关系。注意不能用map< int, vector > 来存储一个machine和一串id,这样就没法找某个id对应哪个machine了。
另外还要 用一个set来存储已经用过的id。
另外再强调map和set的key都是默认从大到小排序的。记得lower_bound()这个函数非常非常有用!map和set都可以用!
class Solution {
public:
int n, k;
map<int, int> shards; //stores id<->machine mapping, multiple ids to one machine
set<int> ids; //stores ids to make sure it is not reused
/*
* @param n: a positive integer
* @param k: a positive integer
* @return: a Solution object
*/
static Solution create(int n, int k) {
Solution sol = Solution();
sol.n = n;
sol.k = k;
return sol;
}
/*
* @param machine_id: An integer
* @return: a list of shard ids
*/
vector<int> addMachine(int machine_id) {
vector<int> nodes;
for (int i = 0; i < k; ++i) {
int id;
do {
id = rand() % n;
} while(shards.find(id) != shards.end());
shards[id] = machine_id;
ids.insert(id);
nodes.push_back(id);
}
sort(nodes.begin(), nodes.end());
return nodes;
}
/*
* @param hashcode: An integer
* @return: A machine id
*/
int getMachineIdByHashCode(int hashcode) {
auto id = shards.lower_bound(hashcode);
if (id == shards.end())
return shards.begin()->second;
else
return id->second;
}
};
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版权声明:本文为CSDN博主「纸上得来终觉浅 绝知此事要躬行」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/roufoo/article/details/81328924
另外还要 用一个set来存储已经用过的id。
另外再强调map和set的key都是默认从大到小排序的。记得lower_bound()这个函数非常非常有用!map和set都可以用!
class Solution {
public:
int n, k;
map<int, int> shards; //stores id<->machine mapping, multiple ids to one machine
set<int> ids; //stores ids to make sure it is not reused
/*
* @param n: a positive integer
* @param k: a positive integer
* @return: a Solution object
*/
static Solution create(int n, int k) {
Solution sol = Solution();
sol.n = n;
sol.k = k;
return sol;
}
/*
* @param machine_id: An integer
* @return: a list of shard ids
*/
vector<int> addMachine(int machine_id) {
vector<int> nodes;
for (int i = 0; i < k; ++i) {
int id;
do {
id = rand() % n;
} while(shards.find(id) != shards.end());
shards[id] = machine_id;
ids.insert(id);
nodes.push_back(id);
}
sort(nodes.begin(), nodes.end());
return nodes;
}
/*
* @param hashcode: An integer
* @return: A machine id
*/
int getMachineIdByHashCode(int hashcode) {
auto id = shards.lower_bound(hashcode);
if (id == shards.end())
return shards.begin()->second;
else
return id->second;
}
};
————————————————
版权声明:本文为CSDN博主「纸上得来终觉浅 绝知此事要躬行」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/roufoo/article/details/81328924
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