LeetCode 241. Different Ways to Add Parentheses

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"
Output: [0, 2]
Explanation: 
((2-1)-1) = 0 
(2-(1-1)) = 2

Example 2:

Input: "2*3-4*5"
Output: [-34, -14, -10, -10, 10]
Explanation: 
(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

这题标签是divide conquer，一开始没什么思路，看了一下花花酱的视频才明白怎么做。。。

关键就是按照operation符号来分割，然后分别递归。。。比如2*3-4*5，就是2   *   3-4*5，然后3-4*5又递归，就是3   -  4*5，如果ans长度为0，就返回int(input),此时Input就是一个数字，另外可以记住每次得到的子串表达式的值，减少运算量，但实际试下来，这样更慢。。不知道为啥。还有个技巧就是pyhton有定义运算function，分别是add,sup,mul,div这个一用舒服很多有没有！！！

from operator import *
class Solution(object):
    def diffWaysToCompute(self, input):
        """
        :type input: str
        :rtype: List[int]
        """
        dic={}
        def helper(input):
            # if input in dic:
            #     return dic[input]
            op={"-":sub,"+":add,"*":mul,"/":div}
            res=[]
            for index,unit in enumerate(input):
                if unit in op:
                    left=helper(input[:index])
                    right=helper(input[index+1:])
                    res.extend([op[unit](a,b) for a in left for b in right])
            if len(res)==0:
                return [int(input)]
            else:
                # dic[input]=res
                return res
        return helper(input)