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]

直接上分治法，至于怎么分治，遍历一遍输入的字符串，把每个+，-，*的左右两边都算好就可以了。需要注意从字符串到数字怎么转换，里面的一些函数需要记住，代码如下：

class Solution {public:    vector<int> diffWaysToCompute(string input)    {        vector<int> result;        for(int i = 0; i < input.size(); i++)        {            if(input[i]=='+'||input[i]=='-'||input[i]=='*')            {                vector<int> left = diffWaysToCompute(input.substr(0,i));//这里刚开始写成input.substr(0,i-1)就WA了                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] == '+')                            result.push_back(left[j]+right[k]);                        else if(input[i] == '-')                            result.push_back(left[j]-right[k]);                        else if(input[i] == '*')                            result.push_back(left[j]*right[k]);                    }                }                }        }        if(result.empty())        {            result.push_back(atoi(input.c_str()));        }        return result;    }};