三合一。描述如何只用一个数组来实现三个栈。
你应该实现push(stackNum, value)、pop(stackNum)、isEmpty(stackNum)、peek(stackNum)方法。stackNum表示栈下标，value表示压入的值。
构造函数会传入一个stackSize参数，代表每个栈的大小。
示例1:
输入：
["TripleInOne", "push", "push", "pop", "pop", "pop", "isEmpty"]
[[1], [0, 1], [0, 2], [0], [0], [0], [0]]
输出：
[null, null, null, 1, -1, -1, true]
说明：当栈为空时`pop, peek`返回-1，当栈满时`push`不压入元素。示例2:
输入：
["TripleInOne", "push", "push", "push", "pop", "pop", "pop", "peek"]
[[2], [0, 1], [0, 2], [0, 3], [0], [0], [0], [0]]
输出：
[null, null, null, null, 2, 1, -1, -1]提示：
- 0 <= stackNum <= 2
Python实现
使用数组直接模拟三个栈即可，初始化为嵌套的空数组。
class TripleInOne:
def __init__(self, stackSize: int):
self.stackSize = stackSize
self.stacks = [[],[],[]]
def push(self, stackNum: int, value: int) -> None:
if len(self.stacks[stackNum])<self.stackSize:
self.stacks[stackNum].append(value)
def pop(self, stackNum: int) -> int:
if self.stacks[stackNum]:
return self.stacks[stackNum].pop()
else:
return -1
def peek(self, stackNum: int) -> int:
if self.stacks[stackNum]:
return self.stacks[stackNum][-1]
return -1
def isEmpty(self, stackNum: int) -> bool:
return len(self.stacks[stackNum])==0
# Your TripleInOne object will be instantiated and called as such:
# obj = TripleInOne(stackSize)
# obj.push(stackNum,value)
# param_2 = obj.pop(stackNum)
# param_3 = obj.peek(stackNum)
# param_4 = obj.isEmpty(stackNum)
