Remove Duplicates from Sorted List[E,Linked List]

https://leetcode.com/problems/remove-duplicates-from-sorted-list/description/

Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

 

Example 1:

Input: head = [1,1,2]
Output: [1,2]

Example 2:

Input: head = [1,1,2,3,3]
Output: [1,2,3]

 

Constraints:

• The number of nodes in the list is in the range [0, 300].
• -100 <= Node.val <= 100
• The list is guaranteed to be sorted in ascending order.

1.C

기본적인건데도 왜이렇게 짜치냐

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* deleteDuplicates(struct ListNode* head) {
    if(!head||!head->next) return head;

    struct ListNode* slow = head;
    struct ListNode* fast = head;

    while(fast){
        while(fast->next && fast->val == fast->next->val){
            fast = fast->next;
        }
        if(slow->val==fast->val) fast=fast->next;
        slow->next = fast;
        slow = fast;
        if(!fast) break;
        fast = fast->next;
    }
    return head;
}

 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* deleteDuplicates(struct ListNode* head) {
    if(!head || !head->next) return head;
    
    struct ListNode* curr = head;
    while(curr && curr->next){
        if(curr->val == curr->next->val){
            curr->next = curr->next->next;}
        else curr = curr->next;
    }
    return head;
}

위와 같이 어찌보면 간단히 할 수 있는 건데;; 에효 좀 헷갈렸다.

 

 

2. C++

비슷한 결로 해결. 중복 delete로 제거까지 포함했다.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(!head || !head->next) return head;

        ListNode* curr = head;
        
        while(curr && curr->next){
            if(curr->val == curr->next->val){
                ListNode* temp = curr->next;
                curr->next = curr->next->next;
                delete temp;
            }
            else curr = curr->next;
        }
        
        return head;
    }
};

 

3. Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next :
            return head
        curr = head
        while curr and curr.next:
            if curr.val == curr.next.val :
                curr.next = curr.next.next
            else :
                curr = curr.next
        return head

 

 

하나 느낀 건, while문을 난발하다보면 코드가 꼬인다는 것이다.

코딩하면서 제일 무서운 while / recursive 구조 ㅠㅠ 언제 친해지냐