题目链接 |
2130. 链表最大孪生和 |
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 pairSum(self, head: Optional[ListNode]) -> int: |
| if head is None: |
| return 0 |
| first = self.halfList(head) |
| second = self.reverseList(first.next) |
| |
| answer = 0 |
| while second: |
| answer = max(answer, head.val + second.val) |
| head = head.next |
| second = second.next |
| 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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