如何按降序对字符串向量进行排序？

Question

I have this following code where there is a vector of strings. Each string is an integer. I want to sort this in a descending order. The regular sort function did not solve my problem. Can someone point out how to do this? I want the output as 345366,38239,029323. I want the leading zero in 029323 as well.

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>

using namespace std;


int main() {
    vector<string> v = {"345366", "029323", "38239"};
    vector<int> temp(v.size());
    for (int idx = 0; idx < v.size(); idx++)
        temp[idx] = stoi(v[idx]);
    sort(temp.begin(), temp.end()));
    cout<<temp[0]<<" "<<temp[1]<<" "<<temp[2];

    return 0;
}

Answer 1

You can use a comparator function like this:

vector<string> v = {"345366", "029323", "38239"};
std::sort(v.begin(), v.end(), [](const std::string &s1, const std::string &s2) -> bool {
    return std::stoi(s1) > std::stoi(s2); 
});
for(auto i : v)
    cout << i << endl;

Check this std::stoi() reference .

Edit: From the comments, it seems std::stoi() is much better than std::atoi() . For converting C++ strings, use std::stoi() . For C strings, std::atoi() will silently fail without generating any error if the string is not convertible to int, whereas std::stoi() will generate an exception, thus is a safer choice as well.

cout << std::atoi("abc") << endl; // runs smoothly
cout << std::stoi("abc") << endl; // creates an 'uncaught exception of type std::invalid_argument: stoi'

However, the results will be the same in this case (will extract the prefix integer part and exit, in case of std::stoi() , if the string doesn't begin with integers, it will create an exception):

cout << std::atoi("999abc12") << endl; // prints 999
cout << std::stoi("999abc12") << endl; // prints 999
cout << std::stoi("abcdef12") << endl; // generates exception

Also see this answer .