从具有特定字符串长度的向量中删除元素

Question

I have a vector that has random strings as elements. I am trying to loop through to find strings that are not a length of 2. If they are not a length of 2 then it should be removed from the vector. For some reason, my code is not removing all of the strings that are not a length of 2.

I have tried individually removing the elements but when it is in a loop it doesn't seem to be working for elements.

This is my output: ho ao lol cd ef wq yo

and expected output should not include the string "lol"

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector <string> numbers = {"ho","ao", "tom", "lol", "cd", "ef", "wq", "hello","yo","sup","boi"};

// loop deletes string elements from vector that don't have length of 2
for(int i=0; i < numbers.size(); i++){
    if(numbers[i].length() != 2){
        numbers.erase(numbers.begin() + i); 
    }
}

// this makes sure the last element in vector gets deleted
if( numbers[numbers.size() - 1].length() != 2  )
    numbers.pop_back();


for(int i=0; i < numbers.size(); i++){
    cout << numbers[i] << endl; // prints out vector elements
}
}

Answer 1

When you erase for the first time (in you situation when you erase "tom") the id of the next elements are decreased, so in next iteration you don't check the length of "lol".

It's not the best solution but you can decrement i in if and your code will work.

if(numbers[i].length() != 2){
    numbers.erase(numbers.begin() + i);
    i--;
}

Answer 2

It's because after you erased "tom" the index has changed.

---

Before erased "tom"

• [0] "ho"
• [1] "ao"
• [2] "tom"
• [3] "lol"
• ...

After

• [0] "ho"
• [1] "ao"
• [2] "lol"
• [3] "cd"

• ...

  You can fix it by decreasing i;

---

if(numbers[i].length() != 2){
    numbers.erase(numbers.begin() + i);
    i--;
}

or you can do this

---

for( auto i = numbers.begin(); i != numbers.end(); )
{
    if( i->length() != 2 ) i = numbers.erase(i); else ++i;
}