Different Ways to Add Parentheses

题目

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*.

Example 1

Input:"2-1-1".

((2-1)-1) = 0
(2-(1-1)) = 2

Output:[0, 2]

Example 2

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]

思路分析

这道题第一次做的时候没有思路，不知道如何下手，后来看了一下divide and conquer的算法，觉得这道题是一个典型的divide and conquer可以实现的题目，因为operator两边的运算可以可以divide成两个子问题，通过递归的方式来实现。得到operator两边的结果之后，再对这两个结果进行conquer合并得到最后的结果。

class Solution(object):
    def diffWaysToCompute(self, input):
        """
        :type input: str
        :rtype: List[int]
        """
        if input.isdigit(): return [int(input)]
        result = []
        for index, char in enumerate(input):
            if char in "+-*":
                leftRes = self.diffWaysToCompute(input[:index])
                rightRes = self.diffWaysToCompute(input[index+1:])
                for num1 in leftRes:
                    for num2 in rightRes:
                        if char == "+": res = num1+num2
                        elif char == "-": res = num1-num2
                        else: res = num1 * num2
                        result.append(res)
        return result