Singly linked list queue class

I have a Queue class that is basically a single linked list. It has 2 remove methods. What would be the cleanest way to avoid duplication of the code? Only the comparisons in methods are different.

int Queue::remove(const string& substring)
{
    // Loop through the queue list
    size_t found;
    found = current_job->description.find(substring)
    if (found != string::npos) {
        // Remove job.
    }
    // Return the amount of removed jobs.   
}

int Queue::remove(int priority)
{
    // Loop through the queue list
    if (current_job->priority == priority) {
        // Remove job.
    }
    // Return the amount of removed jobs.     
}

I thought a template solution as answered here, but in this case, the comparison function has different arguments.

You can abstract out the thing that changes, the comparison expression, into a predicate for a template function.

class PriorityMatch
{
    int priority_;
public:
    explicit PriorityMatch(int priority)
        : priority_(priority) {}

    bool operator()(const Job& job)
    {
        return job->priority == priority_;
    }
};

class DescriptionMatch
{
    std::string substring_;
public:
    explicit DescriptionMatch(const std::string& substring)
        : substring_(substring) {}

    bool operator()(const Job& job)
    {
        return job->description.find(substring_) != std::string::npos;
    }
};

template
int Queue::remove(Predicate predicate)
{
    // Loop through the queue list
    if (predicate(current_job)) {
        // Remove job.
    }
    // Return the amount of removed jobs.     
}

int Queue::remove(const string& substring)
{
    return remove(DescriptionMatch(substring));
}

int Queue::remove(int priority)
{
    return remove(PriorityMatch(substring));
}

Or in C++11:

template
int Queue::remove(Predicate predicate)
{
    // Loop through the queue list
    if (predicate(current_job)) {
        // Remove job.
    }
    // Return the amount of removed jobs.     
}

int Queue::remove(const string& substring)
{
    return remove([](const Job& job){
        return job->description.find(substring) != std::string::npos;
    });
}

int Queue::remove(int priority)
{
    return remove([](const Job& job){
        return job->priority == priority;
    });
}

StackExchange Code Review Q#9900, answer score: 11