Posted on

Description

Submissions

Submission I

#include <bits/stdc++.h>

using namespace std;

vector<string> split_string(string);

// Complete the reverseArray function below.
vector<int> reverseArray(vector<int> a) {
    vector<int> res(a.size());
    reverse_copy(a.begin(), a.end(), res.begin());
    return res;
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    int arr_count;
    cin >> arr_count;
    cin.ignore(numeric_limits<streamsize>::max(), '\n');

    string arr_temp_temp;
    getline(cin, arr_temp_temp);

    vector<string> arr_temp = split_string(arr_temp_temp);

    vector<int> arr(arr_count);

    for (int i = 0; i < arr_count; i++) {
        int arr_item = stoi(arr_temp[i]);

        arr[i] = arr_item;
    }

    vector<int> res = reverseArray(arr);

    for (int i = 0; i < res.size(); i++) {
        fout << res[i];

        if (i != res.size() - 1) {
            fout << " ";
        }
    }

    fout << "\n";

    fout.close();

    return 0;
}

vector<string> split_string(string input_string) {
    string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
        return x == y and x == ' ';
    });

    input_string.erase(new_end, input_string.end());

    while (input_string[input_string.length() - 1] == ' ') {
        input_string.pop_back();
    }

    vector<string> splits;
    char delimiter = ' ';

    size_t i = 0;
    size_t pos = input_string.find(delimiter);

    while (pos != string::npos) {
        splits.push_back(input_string.substr(i, pos - i));

        i = pos + 1;
        pos = input_string.find(delimiter, i);
    }

    splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));

    return splits;
}

Submission II

vector<int> reverseArray(vector<int> a) {
    return vector<int>(a.rbegin(), a.rend());
}

References

  1. https://stackoverflow.com/questions/28148205/copy-a-vector-to-another-vector-in-reverse-order/28148217
  2. http://www.cplusplus.com/reference/algorithm/reverse_copy/

Leave a Reply

Your email address will not be published. Required fields are marked *