61. Rotate List

https://leetcode.com/problems/rotate-list/

Rotate List - LeetCode

 

node 목록과 임의의 정수가 주어졌을 때, 임의의 정수만큼 목록을 rotation 시켜 목록을 반환하는 문제다.

예상과 정답이 다르지 않은 방법이였다.

 

[C++]

/**
 * 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* solve(ListNode* node) {
        ListNode* head = node;
        while (head->next->next) {
            head = head->next;
        }
        
        ListNode* tail = head->next;
        head->next = nullptr;
        tail->next = node;
        return tail;
    }
    
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head || !head->next) {
            return head;
        }
        
        ListNode* backup = head;
        
        int len = 1;
        while (head->next) {
            ++len; head = head->next;
        }
        
        head = backup;
        
        for (int i=0; i<k%len; ++i) {
            head = solve(head);
        }
        
        return head;
    }
};

 

 

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
   
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        if not head or not head.next:
            return head
        
        backup = head
        length = 1
        
        while head.next:
            length += 1
            head = head.next
        
        # set infinite rotatin
        head.next = backup
        head = backup
        
        k = k % length
        
        # rotation 돌수록 가장 tail 쪽 노드부터 head가 되므로 length - k - 1
        # rotate = 1
        # 5 - 1 - 1 = 3
        for _ in range(length - k - 1):
            head = head.next
        
        # anwer = head.next = node(4)
        answer = head.next
        # answer.next = node(3).next = None
        head.next = None
        
        return answer