241. Different Ways to Add Parentheses -------M！！！！

Q:

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]

A:          分而治之， 从每个+-* 拆开

class Solution {
public:
    vector<int> diffWaysToCompute(string input) { // assume no space
        bool foundOp = false;
        vector<int> res;
        for(int i =0;i<input.size();i++){
            if(!isdigit(input[i])){
                foundOp=true;
                auto L=diffWaysToCompute(input.substr(0,i));
                auto R=diffWaysToCompute(input.substr(i+1));
                for(auto s1 : L)
                    for(auto s2 : R)
                        res.push_back(cal(s1, s2, input[i]));
            }
        }
        if(not foundOp)
            res.push_back(stoi(input));
        return res;        
    }
private:
    int cal(int a, int b, char op){
        if(op=='+')
            return a+b;
        if(op=='-')
            return a-b;
        if(op=='*')
            return a*b;
        return 0; // wrong . should never hit
    }
};

Mistakes:

一开始，先split成一个个vector<string> 其实是没有必要的。 以后注意