Bfor(int i = 0; i < list.Count; i++) { if(list.ElementAt(i) % 2 == 0) list.Remove(list.ElementAt(i)); }
Cvar current = list.First; while(current != null) { var next = current.Next; if(current.Value % 2 == 0) list.Remove(current); current = next; }
Dlist.RemoveAll(x => x % 2 == 0);
Step-by-Step Solution
Solution:
Step 1: Understand safe removal during iteration
Removing nodes while iterating requires storing next node before removal to avoid invalid references.
Step 2: Analyze each option
foreach(var node in list) { if(node % 2 == 0) list.Remove(node); } uses foreach which throws error on modification during iteration. var current = list.First; while(current != null) { var next = current.Next; if(current.Value % 2 == 0) list.Remove(current); current = next; } correctly uses a while loop with next node saved. for(int i = 0; i < list.Count; i++) { if(list.ElementAt(i) % 2 == 0) list.Remove(list.ElementAt(i)); } uses ElementAt which is inefficient and unsafe. list.RemoveAll(x => x % 2 == 0); is invalid as LinkedList has no RemoveAll method.
Final Answer:
var current = list.First; while(current != null) { var next = current.Next; if(current.Value % 2 == 0) list.Remove(current); current = next; } -> Option C
Quick Check:
Use while loop with next saved to remove nodes safely [OK]
Quick Trick:Save next node before removal to avoid iteration errors. [OK]
Common Mistakes:
MISTAKES
Modifying list inside foreach causes runtime error
Using RemoveAll which LinkedList does not have
Using ElementAt which is inefficient and unsafe
Master "Collections" in C Sharp (C#)
9 interactive learning modes - each teaches the same concept differently
