Topic Edit: 다음 5개함수 중 가장 잘 쓴 함수라고 생각되는 함수는? 이유는? 틀렸다고 생각되는 함수는? 리스트의 뒤에서 N번째 노드 지운 리스트 리턴하기

This topic has [14] replies, 0 voices, and was last updated 3 years ago by w.

Now Editing “다음 5개함수 중 가장 잘 쓴 함수라고 생각되는 함수는? 이유는? 틀렸다고 생각되는 함수는? 리스트의 뒤에서 N번째 노드 지운 리스트 리턴하기”

Name *

Password *

Topic Title (Maximum Length 80)

`
함수 1.
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode * start = new ListNode();  
        start -> next = head;
        ListNode* fast = start;
        ListNode* slow = start;

for(int i = 1; i <= n; ++i)
            fast = fast->next;
    
        while(fast->next != NULL)
        {
            fast = fast->next;
            slow = slow->next;
        }
        
        slow->next = slow->next->next;
        
        return start->next; 
    }

함수 2.
  ListNode* removeNthFromEnd(ListNode* head, int n) {
    ListNode* slow = head;
    ListNode* fast = head;
        
    while (n--)
      fast = fast->next;

if (fast == nullptr) 
      return head->next;   // return slow->next;  라고 해도 됨.

while (fast->next) {
      slow = slow->next;
      fast = fast->next;
    }
    slow->next = slow->next->next;

return head;   
  }

함수 3.
 void removeNthFromEnd(ListNode* head, int n) {
    ListNode* slow = head;
    ListNode* fast = head;
        
    while (n--)
      fast = fast->next;

if (fast == nullptr) 
       head=head->next;   //  slow->next;  라고 해도 됨.

while (fast->next) {
      slow = slow->next;
      fast = fast->next;
    }
    slow->next = slow->next->next; 
  }

함수 4.
 void removeNthFromEnd(ListNode** head, int n) {
    ListNode* slow = *head;
    ListNode* fast = *head;
        
    while (n--)
      fast = fast->next;

if (fast == nullptr) 
       slow=slow->next;

while (fast->next) {
      slow = slow->next;
      fast = fast->next;
    }
    slow->next = slow->next->next; 
  }

함수 5.
  ListNode* removeNthFromEnd(ListNode* head, int n) {
    ListNode *pre = new ListNode(0, head), *slow = pre, *fast = pre;
    while (fast->next != nullptr && n-- > 0) fast = fast->next;
    while (fast->next != nullptr) {
      fast = fast->next;
      slow = slow->next;
    }
    slow->next = slow->next->next;
    return pre->next;
  }
`

I agree to the terms of service