For loop deletion collection trap in java

First look at the code below:

import java.util.LinkedList;
import java.util.List;
 
public class DeleteCollection {
   
  public static void main(String[] args) {
    List<String> list = new LinkedList<String>();
    list.add("a");
    list.add("b");
    list.add("c");
    list.add("d");
    list.add("e");
    for(int i=0;i<list.size();i++){ //循环删除集合中的元素
      list.remove(i);
    }
    System.out.println("还剩余的元素个数："+list.size());
  }
}

The above code should be correct according to the idea, and the output result should also be 0

Look at the actual output below Result:

Number of remaining elements: 2

You may ask why? Because the size of the collection changes dynamically, when you delete an element, the serial numbers in the elements are rearranged. The second element that should be deleted is now ranked at the position of the first element, but the third element is actually deleted. elements, and so on, the first, third, fifth,... get(i)); can be verified.

The output result after adding the above statement:

Elements to be deleted: a
Elements to be deleted: c
Elements to be deleted:e
Number of remaining elements :2

Solution:

The reason is because the element you want to delete has moved to the front, and the value saved by your i still moves to the back, so if i does not move forward, Go back and go one step forward to delete the element that was originally ranked in the second position but is now in the first position.

Changed core code:

for(int i=0;i<list.size();i++){
    System.out.println("即将删除的元素："+list.get(i));
    list.remove(i);
    i--;
  }

Result:

Element to be deleted:a
Element to be deleted:b
To be deleted Elements to be deleted: c
Elements to be deleted: d
Elements to be deleted: e
Number of remaining elements: 0

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