leetcode 241. Different Ways to Add Parentheses （Python版）

題目：

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are+, - and *.

大意是給定一個運算，求解全部運算序列的解

例如

Input: "2*3-4*5"

(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

算法思路：

將運算轉換成棧過程，這樣就將問題轉換成一個遞歸問題

即每一步分爲兩種走法：

    1）繼續向棧中添加數字與運算符

    2）彈出2個數字1個運算符進行計算

代碼：

class Solution(object):
    def addOperatorAndNumber(self, inputList, stack):
        if len(inputList) < 2:
            return []
        stack.append(inputList[0])
        stack.append(inputList[1])

        inputList = inputList[2:]
        return self.calculate(inputList,stack)

    def calculateTopStack(self, inputList, stack):
        number1 = int(stack.pop())
        operator = stack.pop()
        number2 = int(stack.pop())

        if operator == "+":
            stack.append(number1 + number2)
        elif operator == "-":
            stack.append(number2 - number1)
        elif operator == "*":
            stack.append(number1 * number2)

        return self.calculate(inputList,stack)

    def calculate(self, inputList, stack):
        if len(stack) == 1:
            if len(inputList) == 0:
                return [stack[0]]
            else:
                return self.addOperatorAndNumber(inputList,stack)

        result1 = self.calculateTopStack(inputList[:], stack[:])
        result2 = self.addOperatorAndNumber(inputList[:],stack[:])

        result1.extend(result2)
        return result1

    def diffWaysToCompute(self, input):
        """
        :type input: str
        :rtype: List[int]
        """

        input = input.replace('+'," + ")
        input = input.replace("-"," - ")
        input = input.replace("*"," * ")

        inputList = input.split(" ")
        stack = [int(inputList[0])]
        return  self.calculate(inputList[1:],stack)