题目：原题链接（中等）
标签：设计、栈
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | PUSH/POP = O ( 1 ) ; POPAT = O ( N ) | O ( N ) | 120ms (98.53%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class StackOfPlates:
def __init__(self, cap: int):
self.cap = cap
self.stacks = []
def push(self, val: int) -> None:
if self.cap > 0:
if self.stacks and len(self.stacks[-1]) < self.cap:
self.stacks[-1].append(val)
else:
self.stacks.append([val])
def pop(self) -> int:
if not self.stacks:
return -1
val = self.stacks[-1].pop()
if not self.stacks[-1]:
self.stacks.pop()
return val
def popAt(self, index: int) -> int:
if index >= len(self.stacks):
return -1
val = self.stacks[index].pop()
if not self.stacks[index]:
self.stacks.pop(index)
return val
