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The task is to implement an O(1) Least Recently Used Cache
Here is the question on leetcode
https://leetcode.com/problems/lru-cache/
Here is my solution, while it is O(1) it is not the fastest implementationcould you give some feedback and maybe ideas on how can I optimize this ? Thank you !
#include<unordered_map>
#include<list>
class LRUCache {
// umap<key,<value,listiterator>>
// store the key,value, position in list(iterator) where push_back occurred
private:
unordered_map<int,pair<int,list<int>::iterator>> umap;
list<int> klist;
int cap = -1;
public:
LRUCache(int capacity):cap(capacity){
}
int get(int key) {
// if the key exists in the unordered map
if(umap.count(key)){
// remove it from the old position
klist.erase(umap[key].second);
klist.push_back(key);
list<int>::iterator key_loc = klist.end();
umap[key].second = --key_loc;
return umap[key].first;
}
return -1;
}
void put(int key, int value) {
// if key already exists delete it from the the umap and klist
if(umap.count(key)){
klist.erase(umap[key].second);
umap.erase(key);
}
// if the unordered map is at max capacity
if(umap.size() == cap){
umap.erase(klist.front());
klist.pop_front();
}
// finally update klist and umap
klist.push_back(key);
list<int>::iterator key_loc = klist.end();
umap[key].first = value;
umap[key].second = --key_loc;
return;
}
};
/**
* Your LRUCache object will be instantiated and called as such:
* LRUCache* obj = new LRUCache(capacity);
* int param_1 = obj->get(key);
* obj->put(key,value);
*/
Here's some optimizations that might help:
Take this segment of code from the get function:
if(umap.count(key)){
// remove it from the old position
klist.erase(umap[key].second);
The above will lookup key in the map twice. Once for the count method to see if it exists. Another to invoke the [] operator to fetch its value. Save a few cycles by doing this:
auto itor = umap.find(key);
if (itor != umap.end()) {
// remove it from the old position
klist.erase(itor->second);
In the put function, you do this:
if(umap.count(key)){
klist.erase(umap[key].second);
umap.erase(key);
}
Same thing as get, you can avoid the redundant search through umap. Additionally, there's no reason to invoke umap.erase only to add that same key back into the map a few lines later.
Further, this is also inefficient
umap[key].first = value;
umap[key].second = --key_loc;
Similar to above, redundantly looking up key twice in the map. In the first assignment statement, the key is not in the map, so it default constructs a new value pair thing. The second assignment is doing another lookup in the map.
Let's restructure your put function as follows:
void put(int key, int value) {
auto itor = umap.find(key);
bool reinsert = (itor != umap.end());
// if key already exists delete it from the klist only
if (reinsert) {
klist.erase(umap[key].second);
}
else {
// if the unordered map is at max capacity
if (umap.size() == cap) {
umap.erase(klist.front());
klist.pop_front();
}
}
// finally update klist and umap
klist.push_back(key);
list<int>::iterator key_loc = klist.end();
auto endOfList = --key_loc;
if (reinsert) {
itor->second.first = value;
itor->second.second = endOfList;
}
else {
const pair<int, list<int>::iterator> itempair = { value, endOfList };
umap.emplace(key, itempair);
}
}
That's as far as you can probably go by using std::list. The downside of the list type is that there's no way to move an existing node from the middle to the front (or back) without first removing it and then adding it back. That's a couple of unneeded memory allocations to update the list. Possible alternative is that you just use your own double-linked list type and manually fixup the prev/next pointer yourself.
Here is my solution, while it is O(1) it is not the fastest implementation
could you give some feedback and maybe ideas on how can I optimize this ? Thank you !
Gonna take on selbie's point here:
Every instance of if(umap.count(key)) will search for the key and using umap[key] is the equivalent for the search. You can avoid the double search by assigning an iterator which points to the key by a single std::unordered_map::find() operation.
selbie already gave the code for int get()'s search, here's the one for void put()'s one:
auto it = umap.find(key);
if (it != umap.end())
{
klist.erase(it ->second);
umap.erase(key);
}
Sidecase:
Not applicable for your code as of now due to lack of input and output work, but in case you use std::cin and std::cout, you can disable the synchronization between C and C++ streams, and untie cin from cout as an optimization: (they are tied together by default)
// If your using cin/cout or I/O
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
Working on below problem as an algorithm puzzle. Referred a few similar solutions (and post one of them below), tried and they worked. The question is, for the line "swap(num[i], num[k]);", how do we ensure we could always swap to a number which never tried before (e.g. suppose we swap 1 with 2 in current iteration of the for loop, then it is possible later we swap 2 back with 1 in next iterations of the same for loop of the same level/layer of recursive call)? I have the confusion since we pass num by reference, and it is very possible later (lower level/layer) recursive calls modify content of num, which cause numbers we already evaluated swap back. However, I tried and it works for all of my test cases. Wondering if below solution is 100% correct, or happened to pass my test cases? :)
Here are detailed problem statement and code I am debugging,
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1]
class Solution {
public:
void recursion(vector<int> num, int i, int j, vector<vector<int> > &res) {
if (i == j-1) {
res.push_back(num);
return;
}
for (int k = i; k < j; k++) {
if (i != k && num[i] == num[k]) continue;
swap(num[i], num[k]);
recursion(num, i+1, j, res);
}
}
vector<vector<int> > permuteUnique(vector<int> &num) {
sort(num.begin(), num.end());
vector<vector<int> >res;
recursion(num, 0, num.size(), res);
return res;
}
};
thanks in advance,
Lin
As #notmyfriend said in the comments, num is actually copied each function call.
So now it boils down to:
Of all array values, select one to be the first one and place it there.
That in a loop for each value one time, and then recursively:
Of all values after the first one, select one to be the first and place it there...
...and so on, combined with a check to filter out swaps where nothing changes, ie. filter out duplicates.
If num were a real reference, it won't work anymore (at least not without additional steps).
Eg. 1 1 2 is an easy conterexample, it would give the results:
112, 121, 211, 112, 121
ie. there are duplicates despite the check (and probably there
are examples where some permutations are not generated at all, too).
About the comment:
Per default, every normal function parameter in C++ is copied
(normal = without explicit reference symbol '&' etc.).
Maybe you're thinking of C-style arrays: Essentially, what is passed there is a pointer (to the first value). The pointer is copied, but both original and copied pointer point to the same memory location.
While the purpose of std::vector is (too) to contain an array, the vector itself is a single class object (which contains a pointer to the values somewhere). A class can define itself how it should be copied (with a copy constructor).
Technically, the vector class could implement copying as pointer copying, then it would have the same effect as passing the whole vector as reference; but the C++ creators wanted to keep the copy semantics, ie. that copying a container class should make a real copy with all values duplicated.
For non-copying, there are references already...
Below you can find a solution written in Java. Sorry for not providing a solution in C++, I'm not using it for a long time. But the syntax would be similar.
Solution is using Backtracking (https://en.wikipedia.org/wiki/Backtracking)
Also I'm using hashset to check uniqueness, may be there is a solution which does not use any hashset type data structure, becase my solution is using extra memory in order to provide unique solutions.
Sample input and output;
input : [1, 1, 2]
output : [1, 1, 2]
[1, 2, 1]
[2, 1, 1]
And the solution is;
public class permutation {
public static void main(String[] args) {
permutation p = new permutation();
p.permute(new int[] { 1, 1, 2 });
}
HashSet<String> set = new HashSet<String>();
private void permute(int[] arr) {
set.clear();
this.permute(arr, 0, arr.length - 1);
}
private void permute(int[] arr, int l, int r) {
if (l == r) {
String key = Arrays.toString(arr);
if (set.contains(key))
return;
set.add(key);
System.out.println(key);
} else {
for (int i = l; i <= r; i++) {
swap(arr, l, i);
permute(arr, l + 1, r);
swap(arr, i, l);
}
}
}
private void swap(int[] arr, int l, int r) {
int tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
}
}
I need to update maps of type map<string, vector<int>>. I created this function:
// INPUT: d - original map, key - key to be updated,
// new_val - value that will be appended to d[key]
// OUTPUT: vector<int> which is the updated value of d[key]
vector<int> update_value(map<string, vector<int>> d, string key, int new_val) {
map<string, vector<int>>::iterator it = d.find(key);
vector<int> current_vec;
int prev_len = current_vec.size();
if (it != d.end()) {
current_vec = it->second;
current_vec.push_back(new_val);
return current_vec;
}
assert(prev_len + 1, current_vec.size()); // this fails
return current_vec;
}
I always get the assertion statment failed.
What is the correct method to do this?
Your assert will always fail since the current_vec will always be empty in case the key is not found in the map. I suggest you remove this temporary vector, and you handle the case were the key is not found in another way (e.g. insertion).
You also need to pass your structure d by reference in order for it to get updated
vector<int>& update_value(map<string, vector<int>>& d, string key, int new_val) {
map<string, vector<int>>::iterator it = d.find(key);
if (it != d.end()) {
it->second.push_back(new_val);
}
else {
vector<int> v;
v.push_back(new_val);
d.insert(make_pair(key, v));
return d[key];
}
return it->second;
}
If I got this right, it seems the update will take place only if the key actually exists, as the update code is only present in the the body of the if.
On the other hand, the assert will get checked only when the condition it != d.end() (i.e., the key is present in the map) does not hold (otherwise the final return statement in the if body will cause the end of the function).
As a result, when reaching assert you know current_vec.size() (as thus prev_len) to be 0 and so the assert reduces to assert(1, 0), which is false.
As a side note, since you are passing your map by value, the update you make will not reflect in the original dictionary. Similarly, notice that you are also copying the std::vector out of the map.
The problem is: Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s = "hithere",
dict = ["hi", "there"].
Return true because "hithere" can be segmented as "leet code".
My implementation is as below. This code is ok for normal cases. However, it suffers a lot for input like:
s = "aaaaaaaaaaaaaaaaaaaaaaab", dict = {"aa", "aaaaaa", "aaaaaaaa"}.
I want to memorize the processed substrings, however, I cannot done it right. Any suggestion on how to improve? Thanks a lot!
class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict) {
int len = s.size();
if(len<1) return true;
for(int i(0); i<len; i++) {
string tmp = s.substr(0, i+1);
if((wordDict.find(tmp)!=wordDict.end())
&& (wordBreak(s.substr(i+1), wordDict)) )
return true;
}
return false;
}
};
It's logically a two-step process. Find all dictionary words within the input, consider the found positions (begin/end pairs), and then see if those words cover the whole input.
So you'd get for your example
aa: {0,2}, {1,3}, {2,4}, ... {20,22}
aaaaaa: {0,6}, {1,7}, ... {16,22}
aaaaaaaa: {0,8}, {1,9} ... {14,22}
This is a graph, with nodes 0-23 and a bunch of edges. But node 23 b is entirely unreachable - no incoming edge. This is now a simple graph theory problem
Finding all places where dictionary words occur is pretty easy, if your dictionary is organized as a trie. But even an std::map is usable, thanks to its equal_range method. You have what appears to be an O(N*N) nested loop for begin and end positions, with O(log N) lookup of each word. But you can quickly determine if s.substr(begin,end) is a still a viable prefix, and what dictionary words remain with that prefix.
Also note that you can build the graph lazily. Staring at begin=0 you find edges {0,2}, {0,6} and {0,8}. (And no others). You can now search nodes 2, 6 and 8. You even have a good algorithm - A* - that suggests you try node 8 first (reachable in just 1 edge). Thus, you'll find nodes {8,10}, {8,14} and {8,16} etc. As you see, you'll never need to build the part of the graph that contains {1,3} as it's simply unreachable.
Using graph theory, it's easy to see why your brute-force method breaks down. You arrive at node 8 (aaaaaaaa.aaaaaaaaaaaaaab) repeatedly, and each time search the subgraph from there on.
A further optimization is to run bidirectional A*. This would give you a very fast solution. At the second half of the first step, you look for edges leading to 23, b. As none exist, you immediately know that node {23} is isolated.
In your code, you are not using dynamic programming because you are not remembering the subproblems that you have already solved.
You can enable this remembering, for example, by storing the results based on the starting position of the string s within the original string, or even based on its length (because anyway the strings you are working with are suffixes of the original string, and therefore its length uniquely identifies it). Then, in the beginning of your wordBreak function, just check whether such length has already been processed and, if it has, do not rerun the computations, just return the stored value. Otherwise, run computations and store the result.
Note also that your approach with unordered_set will not allow you to obtain the fastest solution. The fastest solution that I can think of is O(N^2) by storing all the words in a trie (not in a map!) and following this trie as you walk along the given string. This achieves O(1) per loop iteration not counting the recursion call.
Thanks for all the comments. I changed my previous solution to the implementation below. At this point, I didn't explore to optimize on the dictionary, but those insights are very valuable and are very much appreciated.
For the current implementation, do you think it can be further improved? Thanks!
class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict) {
int len = s.size();
if(len<1) return true;
if(wordDict.size()==0) return false;
vector<bool> dq (len+1,false);
dq[0] = true;
for(int i(0); i<len; i++) {// start point
if(dq[i]) {
for(int j(1); j<=len-i; j++) {// length of substring, 1:len
if(!dq[i+j]) {
auto pos = wordDict.find(s.substr(i, j));
dq[i+j] = dq[i+j] || (pos!=wordDict.end());
}
}
}
if(dq[len]) return true;
}
return false;
}
};
Try the following:
class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict)
{
for (auto w : wordDict)
{
auto pos = s.find(w);
if (pos != string::npos)
{
if (wordBreak(s.substr(0, pos), wordDict) &&
wordBreak(s.substr(pos + w.size()), wordDict))
return true;
}
}
return false;
}
};
Essentially one you find a match remove the matching part from the input string and so continue testing on a smaller input.
Is there a way to find a nonexisting key in a map?
I am using std::map<int,myclass>, and I want to automatically generate a key for new items. Items may be deleted from the map in different order from their insertion.
The myclass items may, or may not be identical, so they can not serve as a key by themself.
During the run time of the program, there is no limit to the number of items that are generated and deleted, so I can not use a counter as a key.
An alternative data structure that have the same functionality and performance will do.
Edit
I trying to build a container for my items - such that I can delete/modify items according to their keys, and I can iterate over the items. The key value itself means nothing to me, however, other objects will store those keys for their internal usage.
The reason I can not use incremental counter, is that during the life-span of the program they may be more than 2^32 (or theoretically 2^64) items, however item 0 may theoretically still exist even after all other items are deleted.
It would be nice to ask std::map for the lowest-value non-used key, so i can use it for new items, instead of using a vector or some other extrnal storage for non-used keys.
I'd suggest a combination of counter and queue. When you delete an item from the map, add its key to the queue. The queue then keeps track of the keys that have been deleted from the map so that they can be used again. To get a new key, you first check if the queue is empty. If it isn't, pop the top index off and use it, otherwise use the counter to get the next available key.
Let me see if I understand. What you want to do is
look for a key.
If not present, insert an element.
Items may be deleted.
Keep a counter (wait wait) and a vector. The vector will keep the ids of the deleted items.
When you are about to insert the new element,look for a key in the vector. If vector is not empty, remove the key and use it. If its empty, take one from the counter (counter++).
However, if you neveer remove items from the map, you are just stuck with a counter.
Alternative:
How about using the memory address of the element as a key ?
I would say that for general case, when key can have any type allowed by map, this is not possible. Even ability to say whether some unused key exists requires some knowledge about type.
If we consider situation with int, you can store std::set of contiguous segments of unused keys (since these segments do not overlap, natural ordering can be used - simply compare their starting points). When a new key is needed, you take the first segment, cut off first index and place the rest in the set (if the rest is not empty). When some key is released, you find whether there are neighbour segments in the set (due to set nature it's possible with O(log n) complexity) and perform merging if needed, otherwise simply put [n,n] segment into the set.
in this way you will definitely have the same order of time complexity and order of memory consumption as map has independently on requests history (because number of segments cannot be more than map.size()+1)
something like this:
class TKeyManager
{
public:
TKeyManager()
{
FreeKeys.insert(
std::make_pair(
std::numeric_limits<int>::min(),
std::numeric_limits<int>::max());
}
int AlocateKey()
{
if(FreeKeys.empty())
throw something bad;
const std::pair<int,int> freeSegment=*FreeKeys.begin();
if(freeSegment.second>freeSegment.first)
FreeKeys.insert(std::make_pair(freeSegment.first+1,freeSegment.second));
return freeSegment.first;
}
void ReleaseKey(int key)
{
std:set<std::pair<int,int>>::iterator position=FreeKeys.insert(std::make_pair(key,key)).first;
if(position!=FreeKeys.begin())
{//try to merge with left neighbour
std::set<std::pair<int,int>>::iterator left=position;
--left;
if(left->second+1==key)
{
left->second=key;
FreeKeys.erase(position);
position=left;
}
}
if(position!=--FreeKeys.end())
{//try to merge with right neighbour
std::set<std::pair<int,int>>::iterator right=position;
++right;
if(right->first==key+1)
{
position->second=right->second;
FreeKeys.erase(right);
}
}
}
private:
std::set<std::pair<int,int>> FreeKeys;
};
Is there a way to find a nonexisting
key in a map?
I'm not sure what you mean here. How can you find something that doesn't exist? Do you mean, is there a way to tell if a map does not contain a key?
If that's what you mean, you simply use the find function, and if the key doesn't exist it will return an iterator pointing to end().
if (my_map.find(555) == my_map.end()) { /* do something */ }
You go on to say...
I am using std::map, and
I want to automatically generate a key
for new items. Items may be deleted
from the map in different order from
their insertion. The myclass items may, or may not be identical, so they can not serve as a key by themself.
It's a bit unclear to me what you're trying to accomplish here. It seems your problem is that you want to store instances of myclass in a map, but since you may have duplicate values of myclass, you need some way to generate a unique key. Rather than doing that, why not just use std::multiset<myclass> and just store duplicates? When you look up a particular value of myclass, the multiset will return an iterator to all the instances of myclass which have that value. You'll just need to implement a comparison functor for myclass.
Could you please clarify why you can not use a simple incremental counter as auto-generated key? (increment on insert)? It seems that there's no problem doing that.
Consider, that you decided how to generate non-counter based keys and found that generating them in a bulk is much more effective than generating them one-by-one.
Having this generator proved to be "infinite" and "statefull" (it is your requirement), you can create a second fixed sized container with say 1000 unused keys.
Supply you new entries in map with keys from this container, and return keys back for recycling.
Set some low "threshold" to react on key container reaching low level and refill keys in bulk using "infinite" generator.
The actual posted problem still exists "how to make efficient generator based on non-counter". You may want to have a second look at the "infinity" requirement and check if say 64-bit or 128-bit counter still can satisfy your algorithms for some limited period of time like 1000 years.
use uint64_t as a key type of sequence or even if you think that it will be not enough
struct sequence_key_t {
uint64_t upper;
uint64_t lower;
operator++();
bool operator<()
};
Like:
sequence_key_t global_counter;
std::map<sequence_key_t,myclass> my_map;
my_map.insert(std::make_pair(++global_counter, myclass()));
and you will not have any problems.
Like others I am having difficulty figuring out exactly what you want. It sounds like you want to create an item if it is not found. sdt::map::operator[] ( const key_type& x ) will do this for you.
std::map<int, myclass> Map;
myclass instance1, instance2;
Map[instance1] = 5;
Map[instance2] = 6;
Is this what you are thinking of?
Going along with other answers, I'd suggest a simple counter for generating the ids. If you're worried about being perfectly correct, you could use an arbitrary precision integer for the counter, rather than a built in type. Or something like the following, which will iterate through all possible strings.
void string_increment(std::string& counter)
{
bool carry=true;
for (size_t i=0;i<counter.size();++i)
{
unsigned char original=static_cast<unsigned char>(counter[i]);
if (carry)
{
++counter[i];
}
if (original>static_cast<unsigned char>(counter[i]))
{
carry=true;
}
else
{
carry=false;
}
}
if (carry)
{
counter.push_back(0);
}
}
e.g. so that you have:
std::string counter; // empty string
string_increment(counter); // now counter=="\x00"
string_increment(counter); // now counter=="\x01"
...
string_increment(counter); // now counter=="\xFF"
string_increment(counter); // now counter=="\x00\x00"
string_increment(counter); // now counter=="\x01\x00"
...
string_increment(counter); // now counter=="\xFF\x00"
string_increment(counter); // now counter=="\x00\x01"
string_increment(counter); // now counter=="\x01\x01"
...
string_increment(counter); // now counter=="\xFF\xFF"
string_increment(counter); // now counter=="\x00\x00\x00"
string_increment(counter); // now counter=="\x01\x00\x00"
// etc..
Another option, if the working set actually in the map is small enough would be to use an incrementing key, then re-generate the keys when the counter is about to wrap. This solution would only require temporary extra storage. The hash table performance would be unchanged, and the key generation would just be an if and an increment.
The number of items in the current working set would really determine if this approach is viable or not.
I loved Jon Benedicto's and Tom's answer very much. To be fair, the other answers that only used counters may have been the starting point.
Problem with only using counters
You always have to increment higher and higher; never trying to fill the empty gaps.
Once you run out of numbers and wrap around, you have to do log(n) iterations to find unused keys.
Problem with the queue for holding used keys
It is easy to imagine lots and lots of used keys being stored in this queue.
My Improvement to queues!
Rather than storing single used keys in the queue; we store ranges of unused keys.
Interface
using Key = wchar_t; //In my case
struct Range
{
Key first;
Key last;
size_t size() { return last - first + 1; }
};
bool operator< (const Range&,const Range&);
bool operator< (const Range&,Key);
bool operator< (Key,const Range&);
struct KeyQueue__
{
public:
virtual void addKey(Key)=0;
virtual Key getUniqueKey()=0;
virtual bool shouldMorph()=0;
protected:
Key counter = 0;
friend class Morph;
};
struct KeyQueue : KeyQueue__
{
public:
void addKey(Key)override;
Key getUniqueKey()override;
bool shouldMorph()override;
private:
std::vector<Key> pool;
friend class Morph;
};
struct RangeKeyQueue : KeyQueue__
{
public:
void addKey(Key)override;
Key getUniqueKey()override;
bool shouldMorph()override;
private:
boost::container::flat_set<Range,std::less<>> pool;
friend class Morph;
};
void morph(KeyQueue__*);
struct Morph
{
static void morph(const KeyQueue &from,RangeKeyQueue &to);
static void morph(const RangeKeyQueue &from,KeyQueue &to);
};
Implementation
Note: Keys being added are assumed to be key not found in queue
// Assumes that Range is valid. first <= last
// Assumes that Ranges do not overlap
bool operator< (const Range &l,const Range &r)
{
return l.first < r.first;
}
// Assumes that Range is valid. first <= last
bool operator< (const Range &l,Key r)
{
int diff_1 = l.first - r;
int diff_2 = l.last - r;
return diff_1 < -1 && diff_2 < -1;
}
// Assumes that Range is valid. first <= last
bool operator< (Key l,const Range &r)
{
int diff = l - r.first;
return diff < -1;
}
void KeyQueue::addKey(Key key)
{
if(counter - 1 == key) counter = key;
else pool.push_back(key);
}
Key KeyQueue::getUniqueKey()
{
if(pool.empty()) return counter++;
else
{
Key key = pool.back();
pool.pop_back();
return key;
}
}
bool KeyQueue::shouldMorph()
{
return pool.size() > 10;
}
void RangeKeyQueue::addKey(Key key)
{
if(counter - 1 == key) counter = key;
else
{
auto elem = pool.find(key);
if(elem == pool.end()) pool.insert({key,key});
else // Expand existing range
{
Range &range = (Range&)*elem;
// Note at this point, key is 1 value less or greater than range
if(range.first > key) range.first = key;
else range.last = key;
}
}
}
Key RangeKeyQueue::getUniqueKey()
{
if(pool.empty()) return counter++;
else
{
Range &range = (Range&)*pool.begin();
Key key = range.first++;
if(range.first > range.last) // exhausted all keys in range
pool.erase(pool.begin());
return key;
}
}
bool RangeKeyQueue::shouldMorph()
{
return pool.size() == 0 || pool.size() == 1 && pool.begin()->size() < 4;
}
void morph(KeyQueue__ *obj)
{
if(KeyQueue *queue = dynamic_cast<KeyQueue*>(obj))
{
RangeKeyQueue *new_queue = new RangeKeyQueue();
Morph::morph(*queue,*new_queue);
obj = new_queue;
}
else if(RangeKeyQueue *queue = dynamic_cast<RangeKeyQueue*>(obj))
{
KeyQueue *new_queue = new KeyQueue();
Morph::morph(*queue,*new_queue);
obj = new_queue;
}
}
void Morph::morph(const KeyQueue &from,RangeKeyQueue &to)
{
to.counter = from.counter;
for(Key key : from.pool) to.addKey(key);
}
void Morph::morph(const RangeKeyQueue &from,KeyQueue &to)
{
to.counter = from.counter;
for(Range range : from.pool)
while(range.first <= range.last)
to.addKey(range.first++);
}
Usage:
int main()
{
std::vector<Key> keys;
KeyQueue__ *keyQueue = new KeyQueue();
srand(time(NULL));
bool insertKey = true;
for(int i=0; i < 1000; ++i)
{
if(insertKey)
{
Key key = keyQueue->getUniqueKey();
keys.push_back(key);
}
else
{
int index = rand() % keys.size();
Key key = keys[index];
keys.erase(keys.begin()+index);
keyQueue->addKey(key);
}
if(keyQueue->shouldMorph())
{
morph(keyQueue);
}
insertKey = rand() % 3; // more chances of insert
}
}