题目链接 |
234. 回文链表 |
思路 |
链表综合题:快慢指针 + 指针翻转 |
题解链接 |
官方题解 |
关键点 |
while fast.next and fast.next.next: ... |
时间复杂度 |
|
空间复杂度 |
|
代码实现:
| class Solution: |
| def isPalindrome(self, head: Optional[ListNode]) -> bool: |
| if head is None: |
| return True |
| first = self.halfList(head) |
| second = self.reverseList(first.next) |
| |
| answer = True |
| cur_first, cur_second = head, second |
| while answer and cur_second: |
| if cur_first.val != cur_second.val: |
| answer = False |
| break |
| cur_first = cur_first.next |
| cur_second = cur_second.next |
| first.next = self.reverseList(second) |
| return answer |
| |
| def halfList(self, head): |
| slow = fast = head |
| while fast.next and fast.next.next: |
| fast = fast.next.next |
| slow = slow.next |
| return slow |
| |
| def reverseList(self, head): |
| previous = None |
| current = head |
| while current: |
| next_node = current.next |
| current.next = previous |
| previous = current |
| current = next_node |
| return previous |
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