### Using adjacent_find with an iterator ?

Ok I want to be able to not only return when `numbers` finds a match but i want to know how many times they match. In the case below you will see that it will tell me "10 is a match of numbers" but if `numbers;` contained 10 , 10 , 10 how can I get it to tell me "there are 3 matches" or as of now tell me "there are 2 matches"

 ``1234567891011121314151617`` ``````#include #include #include int main() { std::vector numbers; //Contains 10,20,10 std::vector::iterator it; //Search numbers for matches it = std::adjacent_find (numbers.begin(), numbers.end()); //Tell me if they matched if (it!=numbers.end()){ std::cout << *it << " Is a match of numbers" '\n'; } }``````

I think this is the easiest way of trying to find how many matches there are, also open to any other ideas!
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Could you clarify further? Are you looking for the length of the continuous run of equal numbers in an otherwise unsorted sequence (e.g. "70, -1, 10, 10, 10, 20" gives '3') or for the length of a continuous run in a sorted sequence ("-1, 10, 10, 10, 20, 70" gives '3'), or for a total number of matches in an unsorted sequene ("10, -1, 70, 20, 10, 10" gives '3')?
Use of adjacent_find implies the first, but the comment in code ("10, 20, 10") implies the last
Sorry I was just using a basic example! `numbers` is sorted by size order so the vector would show 10,10,20 , all I want to know is how many duplicates there are inside the vector I sorted them in size order so I could use adjacent find but I am not sure if that is the best way? its working but it doesnt tell me how many duplicates it contains
 ``1234567891011121314151617181920212223`` ``````#include #include #include int main() { typedef std::vector vec_type; vec_type numbers { 10, 10, 20, 20, 20, 30, 30, 30, 40, 30, 10, 10 }; auto it = std::begin(numbers); while ((it = std::adjacent_find(it, std::end(numbers))) != std::end(numbers)) { auto value = *it; unsigned matches = 1; while (++it != std::end(numbers) && *it == value) ++matches; std::cout << value << ": " << matches << " matches.\n" ; } }``````

http://ideone.com/aZz6vq
if it's sorted, you can find the end of the range in logarithmic time (in the size of the container) using upper_bound:

 ``1234567`` `````` //Tell me if they matched if (it!=numbers.end()){ std::cout << *it << " Is a match of " << std::upper_bound(it, numbers.end(), *it) - it << " numbers\n"; } }``````

or run down to the end of it with find_if with linear time in the length of the subsequence
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Guys thanks, Cire that works perfectly , its exactly what I needed ! now I can continue with my code but I am going to be spending a while breaking down exactly how it works (really quite new to vectors+iterators) , thanks again really appreciate it :)
Actually just noticed a problem with that Cire.

If the first number is lower than the second two numbers following it , it return "1 match" and shows the low number. Example:

numbers = 5,15,15 . I get a cout of "5: 1 matches". Rather than "15: two matches"

but...

numbers = 5,5,15. I get the correct cout of "5: 2 matches"

so anytime the pair of numbers is greater than the lower single digit , it displays incorrectly using the wrong values, any ideas?
Strange. adjacent_find should return iterator to first 15 in the {5, 15, 15}.

Cubbi wrote:
or run down to the end of it with find_if with linear time in the length of the subsequence

There is also the std::count
Yeh I am not really sure how to fix it and I have never used std::count , any suggestions :( ?
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