｜ 用一個數(shù)組實現(xiàn)三個棧

三合一。描述如何只用一個數(shù)組來實現(xiàn)三個棧。你應該實現(xiàn)push(stackNum, value)、pop(stackNum)、isEmpty(stackNum)、peek(stackNum)方法。stackNum表示棧下標，value表示壓入的值。構造函數(shù)會傳入一個stackSize參數(shù)，代表每個棧的大小。示例1:輸入： ["TripleInOne", "push", "push", "pop", "pop", "pop", "isEmpty"] [[1], [0, 1], [0, 2], [0], [0], [0], [0]]輸出： [null, null, null, 1, -1, -1, true] 說明：當棧為空時`pop, peek`返回-1，當棧滿時`push`不壓入元素。 示例2:輸入： ["TripleInOne", "push", "push", "push", "pop", "pop", "pop", "peek"] [[2], [0, 1], [0, 2], [0, 3], [0], [0], [0], [0]]輸出： [null, null, null, null, 2, 1, -1, -1]

｜ 題解

class TripleInOne:def __init__(self, stackSize: int): # stackSize是棧的大小self.stack = [0 for i in range(stackSize * 3)] # 初始化三倍stackSize的列表存放3個棧self.ptr1 = 0 # 第一個棧的指針，起始位置是列表的0位置self.ptr2 = 0 + stackSize # 第二個棧的指針，起始位置是列表的stackSize位置self.ptr3 = 0 + 2 * stackSize # 第三個棧的指針，起始位置是列表的2*stackSize位置self.stackSize = stackSize # 每一個棧的大小def push(self, stackNum: int, value: int) -> None:if stackNum == 0 and 0 <= self.ptr1 < self.stackSize: # 往第一個棧加入元素self.stack[self.ptr1] = value self.ptr1 += 1elif stackNum == 1 and self.stackSize <= self.ptr2 < 2 * self.stackSize: # 往第二個棧加入元素self.stack[self.ptr2] = valueself.ptr2 += 1elif stackNum == 2 and 2 * self.stackSize <= self.ptr3 < 3 * self.stackSize: # 往第三個棧加入元素self.stack[self.ptr3] = valueself.ptr3 += 1def pop(self, stackNum: int) -> int: # pop() 返回棧頂元素，并在進程中刪除它if stackNum == 0 and 0 < self.ptr1 <= self.stackSize: # 從第一個棧刪除一個元素self.ptr1 -= 1return self.stack[self.ptr1]elif stackNum == 1 and self.stackSize < self.ptr2 <= 2 * self.stackSize: # 從第二個棧刪除一個元素self.ptr2 -= 1return self.stack[self.ptr2]elif stackNum == 2 and 2 * self.stackSize < self.ptr3 <= 3 * self.stackSize: # 從第三個棧刪除一個元素self.ptr3 -= 1return self.stack[self.ptr3]return -1def peek(self, stackNum: int) -> int: # peek() 返回棧頂元素，但不在堆棧中刪除它if stackNum == 0 and 0 < self.ptr1 <= self.stackSize: # 返回第一個棧的棧頂元素return self.stack[self.ptr1-1]elif stackNum == 1 and self.stackSize < self.ptr2 <= 2 * self.stackSize: # 返回第二個棧的棧頂元素return self.stack[self.ptr2-1]elif stackNum == 2 and 2 * self.stackSize < self.ptr3 <= 3 * self.stackSize: # 返回第三個棧的棧頂元素return self.stack[self.ptr3-1]return -1def isEmpty(self, stackNum: int) -> bool: if stackNum == 0 and self.ptr1 == 0:return Trueelif stackNum == 1 and self.ptr2 == 0 + self.stackSize:return Trueelif stackNum == 2 and self.ptr3 == 0 + 2 * self.stackSize:return Truereturn 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)

總結

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