I have this map: map<int, int > items
.
Given a key, I want that this map returns the item corresponding to the key if it is present, otherwise the map returns the item with key immediately less than the given key.
For example, if I have:
items[0]=0;
items[6]=10;
items[15]=18;
items[20]=22;
then for key=15, I want that the map returns item with value 18, otherwise for key=9, I want that map returns item with value 10.
I haven't found a function for this case. But I tried in this way:
itlow=items.lower_bound(key);
if(!items.count(key))
itlow--;
return itlow->second;
This works as I want, entering in the map a min key items[0]=0
for default, but I know that itlow--;
is not good programming. What can I do?
You just need to check if your itlow
is already items.begin()
. If it is, there's no such element in the map:
itlow=items.lower_bound(key);
if(itlow->first == key)
return itlow->second;
else if(itlow != items.begin())
itlow--;
return itlow->second;
else
throw some_exception();
Instead of throwing exception, you may return iterator, and then you can return items.end()
if no such element is found.
#include <iostream>
#include <map>
using namespace std;
map<int, int>::const_iterator find(const map<int, int> &items, int value)
{
auto itlow = items.lower_bound(value);
if(itlow->first == value)
return itlow;
else if(itlow != items.cbegin())
return --itlow;
else
return items.cend();
}
int main()
{
map<int, int> items;
items[2]=0;
items[6]=10;
items[15]=18;
items[20]=22;
auto i = find(items, 0);
if(i != items.cend())
{
cout << i->second << endl;
}
i = find(items, 15);
if(i != items.cend())
{
cout << i->second << endl;
}
i = find(items, 9);
if(i != items.cend())
{
cout << i->second << endl;
}
}