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”.

((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]

这个题目确实不会做，学习参考别人的代码。
自己还是不太擅长这种递归的题目哎。

class Solution {public:    vector<int> diffWaysToCompute(string input) {        vector<int>v;        for(int i=0;i<input.size();i++){            char t=input[i];            if(t=='+'||t=='-'||t=='*'){            vector<int> lv=diffWaysToCompute(input.substr(0,i));            vector<int> rv=diffWaysToCompute(input.substr(i+1));            for(int j=0;j<lv.size();j++){                for(int k=0;k<rv.size();k++){                    if(t=='+')                        v.push_back(lv[j]+rv[k]);                    else if(t=='-')                        v.push_back(lv[j]-rv[k]);                    else if(t=='*')                        v.push_back(lv[j]*rv[k]);                }            }         }        }        if(v.empty())            v.push_back(atoi(input.c_str()));        return v;    }};