「世界で闘うプログラミング力を鍛える本」Pythonコード解答例 - 3.1 3つのスタック

###CHAP3. スタックとキュー

3つのスタック

#Pythonコード解答例


class FixedMultiStack:

    def __init__(self, stackSize):
        numberOfStacks = 3
        self.stackCapacity = stackSize
        self.values = [0] * (stackSize * numberOfStacks) 
        self.sizes = [0] * numberOfStacks

    def indexOfTop(self,stackNum):
        offset = stackNum * self.stackCapacity
        size = self.sizes[stackNum]
        return offset + size - 1

    def isFull(self,stackNum):
        return self.sizes[stackNum] == self.stackCapacity

    def push(self,stackNum,value):
        if self.isFull(stackNum):
            raise Exception("FullStackException")
        self.sizes[stackNum] = self.sizes[stackNum] + 1
        self.values[self.indexOfTop(stackNum)] = value

    def isEmpty(self,stackNum):
        return self.sizes[stackNum] == 0

    def pop(self,stackNum):
        if self.isEmpty(stackNum):
            raise Exception("EmptyStackException")
        topIndex = self.indexOfTop(stackNum)
        value = self.values[topIndex]
        self.values[topIndex] = 0
        self.sizes[stackNum] = self.sizes[stackNum] - 1
        return value

    def peek(self,stackNum):
        if self.isEmpty(stackNum):
            raise Exception("EmptyStackException")
        return self.values[self.indexOfTop(stackNum)]

FMS = FixedMultiStack(5)

print(FMS.values)

FMS.push(0,1)
FMS.push(1,6)
FMS.push(2,11)

print(FMS.values)

FMS.push(0,2)
FMS.push(0,3)
FMS.push(0,4)
FMS.push(0,5)

print(FMS.values)

FMS.pop(0)
FMS.pop(1)

print(FMS.values)

print(FMS.peek(0))
print(FMS.peek(2))