美团暑实面经 已offer

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cpu角度 有点过于刁难，过不了我直接你手撕一个汇编代码出来我看看

class Solution: def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]: slow = fast = head while fast and fast.next: slow = slow.next fast = fast.next.next return slow def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: pre, cur = None, head while cur: nxt = cur.next cur.next = pre pre = cur cur = nxt return pre def reorderList(self, head: Optional[ListNode]) -> None: mid = self.middleNode(head) head2 = self.reverseList(mid) while head2.next: nxt = head.next nxt2 = head2.next head.next = head2 head2.next = nxt head = nxt head2 = nxt2

这是经典的博弈论取硬币问题，核心策略就是控制“5的倍数”这个关键点哦～具体操作很简单： 1. 先手第一步：先取4枚硬币，让剩余硬币数变成60（60是5的倍数）。 2. 后续每轮：对手取x枚（1≤x≤4），你就取“5-x”枚。这样每轮结束后，剩余硬币数始终保持是5的倍数。 3. 最后阶段：当剩余5枚硬币时，对手不管取1-4枚中的多少，你都能取走剩下的所有硬币，赢！ 比如对手取2枚，你就取3枚；对手取4枚，你就取1枚，始终把总数控制在5的倍数上～这样一步步就能把最后一枚硬币稳稳拿到手啦～

因为联合索引是按最左列排序的，跳过最左列会导致索引失效，数据库只能全表扫描

class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next def addTwoNumbers(l1: ListNode, l2: ListNode) -> ListNode: dummy = ListNode() curr = dummy carry = 0 while l1 or l2 or carry: sum_val = carry if l1: sum_val += l1.val l1 = l1.next if l2: sum_val += l2.val l2 = l2.next carry, val = divmod(sum_val, 10) curr.next = ListNode(val) curr = curr.next return dummy.next

class MyQueue: def __init__(self): self.stack_in = [] self.stack_out = [] def push(self, x): self.stack_in.append(x) def pop(self): if not self.stack_out: while self.stack_in: self.stack_out.append(self.stack_in.pop()) return self.stack_out.pop() def peek(self): if not self.stack_out: while self.stack_in: self.stack_out.append(self.stack_in.pop()) return self.stack_out[-1] def empty(self): return not self.stack_in and not self.stack_out

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