程序员面试金典 面试题 03.01. 三合一

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]

解答 python3

class TripleInOne:

    def __init__(self, stackSize: int):
        self.stack=[0 for i in range (3*stackSize)];
        self.top1=0;
        self.top2=0+stackSize;
        self.top3=0+2*stackSize;
        self.stacksize=stackSize;
        


    def push(self, stackNum: int, value: int) -> None:
        if stackNum==0 and 0<=self.top1<self.stacksize:
            self.stack[self.top1]=value;
            self.top1=self.top1+1;
        elif stackNum==1 and self.stacksize<=self.top2<2*self.stacksize:
            self.stack[self.top2]=value;
            self.top2=self.top2+1;
        elif stackNum==2 and 2*self.stacksize<=self.top3<3*self.stacksize:
            self.stack[self.top3]=value;
            self.top3=self.top3+1;

    def pop(self, stackNum: int) -> int:
        if stackNum==0 and 0<self.top1<=self.stacksize:
            
            self.top1=self.top1-1;
            return self.stack[self.top1];
        elif stackNum==1 and self.stacksize<self.top2<=2*self.stacksize:
            
            self.top2=self.top2-1;
            return self.stack[self.top2];
        elif stackNum==2 and 2*self.stacksize<self.top3<=3*self.stacksize:
            
            self.top3=self.top3-1;
            return self.stack[self.top3];
        return -1;

    def peek(self, stackNum: int) -> int:
        if stackNum==0 and 0<self.top1<=self.stacksize:
            return self.stack[self.top1-1];
        elif stackNum==1 and self.stacksize<self.top2<=2*self.stacksize:
            return self.stack[self.top2-1];
        elif stackNum==2 and 2*self.stacksize<self.top3<=3*self.stacksize:
            return self.stack[self.top3-1];
        return -1;
           

    def isEmpty(self, stackNum: int) -> bool:
        if stackNum==0 and 0==self.top1:
            return True;
        elif stackNum==1 and self.stacksize==self.top2:
            return True;
        elif stackNum==2 and 2*self.stacksize==self.top3:
           return True;
        return False;


# 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)

题意也太难理解了！