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".


((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]

思路是 divide and conquer 就是拆分成两个部分，然后在组装在一起

class Solution {public:    vector<int> diffWaysToCompute(string input)     {        vector<int> ret;        for(int i = 0; i < input.size(); i++)        {            if (input[i]=='+' || input[i]=='-' || input[i]=='*')            {                vector<int> left = diffWaysToCompute(input.substr(0, i));                vector<int> right = diffWaysToCompute(input.substr(i + 1));                for (int j = 0; j < left.size(); j++)                {                    for (int k = 0; k < right.size(); k++)                    {                        if (input[i]=='+')                            ret.push_back(left[j] + right[k]);                        else if (input[i]=='-')                            ret.push_back(left[j] - right[k]);                        else if (input[i]=='*')                            ret.push_back(left[j] * right[k]);                    }                }            }        }        if(ret.empty())            ret.push_back(atoi(input.c_str()));        return ret;    }};