LeetCode 솔루션 분류
[8/23] 234. Palindrome Linked List
본문
Easy
10588624Add to ListShareGiven the head of a singly linked list, return true if it is a palindrome.
Example 1:
Input: head = [1,2,2,1] Output: true
Example 2:
Input: head = [1,2] Output: false
Constraints:
- The number of nodes in the list is in the range
[1, 105]. 0 <= Node.val <= 9
Follow up: Could you do it in
O(n) time and O(1) space?Accepted
1,113,887
Submissions
2,317,041
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학부유학생님의 댓글
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Runtime: 1549 ms, faster than 15.03% of Python3 online submissions for Palindrome Linked List.
Memory Usage: 39.2 MB, less than 65.33% of Python3 online submissions for Palindrome Linked List.
Memory Usage: 39.2 MB, less than 65.33% of Python3 online submissions for Palindrome Linked List.
class Solution:
def isPalindrome(self, head: Optional[ListNode]) -> bool:
if not head or not head.next: return True
slow = fast = head
while fast and fast.next:
fast = fast.next.next
slow = slow.next
prev = None
while slow:
nxt = slow.next
slow.next = prev
prev = slow
slow = nxt
first, second = head, prev
while second:
if second.val != first.val: return False
first, second = first.next, second.next
return True
# 1 2 3 2 1
# s
# f
# s f
# s. f
# 1 2 2 1
# s
# f
# s f
# s. f재민재민님의 댓글
- 익명
- 작성일
Runtime: 294 ms, faster than 75.01% of C++ online submissions for Palindrome Linked List.
Memory Usage: 128.3 MB, less than 14.42% of C++ online submissions for Palindrome Linked List.
Memory Usage: 128.3 MB, less than 14.42% of C++ online submissions for Palindrome Linked List.
class Solution {
public:
bool isPalindrome(ListNode* head) {
vector<int> v;
while(head) {
v.push_back(head->val);
head = head->next;
}
for(int i = 0; i < v.size()/2; i++)
if(v[i] != v[v.size()-1-i])
return false;
return true;
}
};
