从向量中删除最后一个元素直到条件

Question

I ran into a problem where i need to delete the last elements of a vector until a certain condition is met (for sake of this example let it be the element is not zero)

I wrote this code, it does the trick -

auto next = vec.rbegin();
while (next != vec.rend())
{
    auto current = next++;
    if (*current == 0)
        vec.pop_back();
    else
        break;
}

But i would much rather find an stl algorithm that i can use (i can use find_if and then erase, but i'd like to loop once through the elements that i remove...)

Also, i'm afraid i may be invoking some UB here, should i be worried?

Answer 1

Your code can be simplier:

while( !vec.empty() && vec.back() == 0 ) 
    vec.pop_back();

Using std::remove or std::remove_if would remove all elements based by criteria, so you should use std::find_if as Vlad provided in his answer.

Answer 2

Here is an example. It uses the general idiom for erasing vectors

v.erase( std::remove( /*...*/ ), v.end() )


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

int main() 
{
    std::vector<int> v = { 1, 2, 3, 4, 5, 0, 0, 0 };

    v.erase( 
        std::find_if( v.rbegin(), v.rend(), 
        []( int x ) { return x != 0; } ).base(), v.end() );

    for ( int x : v ) std::cout << x << ' ';
    std::cout << std::endl;

    return 0;
}

The output is

1 2 3 4 5