std::unique

From Cppreference

Removes all consecutive duplicate elements from the range [first, last). Removing is done by shifting the range when needed in such a way that elements to be erased are overwritten. Only the first element in each group of equal elements is left. The elements between the old and the new end or the range are left intact. The first version uses operator== to compare the elements, the second version uses the given binary predicate p.

[edit] Parameters

[edit] Return value

forward iterator to the new end of the range

[edit] Equivalent function

template<class ForwardIterator>
ForwardIterator unique(ForwardIterator first, ForwardIterator last)
{
    ForwardIterator result = first;
    while (++first != last) {
        if (!(*result == *first)) {
            *(++result) = *first;
        }
    }
    return ++result;
}

template<class ForwardIterator, class BinaryPredicate>
ForwardIterator unique(ForwardIterator first, ForwardIterator last, 
                       BinaryPredicate p)
{
    ForwardIterator result = first;
    while (++first != last) {
        if (!p(*result, *first)) {
            *(++result) = *first;
        }
    }
    return ++result;
}

[edit] Example

The following code removes all consecutive equivalent elements from a vector of integers.

#include <iostream>
#include <algorithm>
#include <vector>
 
int main() 
{
    std::vector<int> v{1, 2, 2, 2, 3, 3, 2, 2, 1};
    std::vector<int>::iterator last;
 
    last = std::unique(v.begin(), v.end()); // 1 2 3 2 1 3 2 2 1
                                            //           ^
    for (std::vector<int>::iterator it = v.begin(); it != last; ++it) {
        std::cout << *it << " ";
    }
    std::cout << "\n";
}

Output:

1 2 3 2 1

[edit] Complexity

linear in the distance between first and last

[edit] See also